Answer:
B - spring scale and meterstick
Explanation:
The work done by the force of gravity as the object falls is equal to the potential energy of the object.
We know that the gravitational potential energy(PE) of an object is obtained by;
PE =mgΔh
Where;
m = mass of the object
g = acceleration due to gravity
h= height of the object
The spring scale is needed to measure the mass of the object while the meterstick measures the Δh in meters.
Answer:
B
Explanation:
w= fd
f: spring scale
d: meterstick
Drop a comment if you have any questions! :)
An 80kg astronaut traveled to the moon, where gravity is one-sixth (116) as
strong as Earth's gravity. What is the astronaut's weight on the moon?
F=mg.g=(1/6)(9.8m/s?)
Answer:
Wmoon = 131 [N]
Explanation:
We know that the weight of a body is equal to the product of mass by gravitational acceleration.
Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.
[tex]w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N][/tex]
a plane flies from Addis Ababa to Gamble is 400km. in 2 hrs, then straight back to Gamma is 200km. in 1hrs what is the average speed, average velocity, total distance and total displacement
Answer:
a) Average speed = 200 Km/hr
b) Average velocity = 0 m/s
c) Total distance = 800 Km
d) Total displacement = 0 Km
Explanation:
a) Let the speed of the plane from Addis Ababa to Gamble be represented by [tex]S_{1}[/tex], and from Gamble to Addis Ababa by [tex]S_{2}[/tex].
Average speed = [tex]\frac{S_{1} + S_{2} }{2}[/tex]
[tex]S_{1}[/tex] = [tex]\frac{distance}{time}[/tex] = [tex]\frac{400}{2}[/tex]
= 200 Km/hr
[tex]S_{2}[/tex] = [tex]\frac{distance}{time}[/tex] = [tex]\frac{200}{1}[/tex]
= 200 Km/hr
Average speed = [tex]\frac{200 + 200}{2}[/tex]
= 200 Km/hr
b) Velocity = [tex]\frac{displacement}{time}[/tex]
Average velocity = [tex]\frac{total displacement}{total time taken}[/tex]
Since the plane flies from Addis Ababa to Gamble, then back to Addis Ababa, its total displacement is zero.
So that,
Average velocity = 0 m/s
c) Total distance = 400 Km + 400 Km
= 800 Km
d) Total displacement = 400 Km + (-400 Km)
= 0 Km
When Coach Kwan notices that a player is getting tired, she takes out the tired player and substitutes a fresh player.
Which type of chemical reaction does this best model?
Answer:
is replacement
Explanation:
Jane climbs the stairs to the first floor all by herself in a certain time. If the next time she rides the elevator to the first floor then which of the following statements is true?
The work done in both cases are different
the time taken in both cases is the same
the power in both cases is the same.
the work done in both cases is the same.
Answer:
the power in both cases is the same.
Explanation:
hope helps you
thanksss
A small steel ball falls from rest through a distance of 3m. When calculating the time of fall, air resistance can be ignored because
Answer:
First, let's write the movement equations for this ball.
The only force acting on the ball will be the gravitational acceleration (because we ignore the air resistance) then the acceleration equation is:
a(t) = -9.8m/s^2
Where the minus sign is because the ball is falling down.
To get the velocity of the ball, we need to integrate over time to get:
v(t) = -(9.8m/s^2)*t + v0
Where v0 is the initial velocity of the ball. Because it falls from rest, we can conclude that the initial velocity is 0 m/s, then the velocity equation is:
v(t) = -(9.8m/s^2)*t
For the position equation we need to integrate again, here we get:
p(t) = -(1/2)*(9.8m/s^2)*t^2 + p0
Where p0 is the initial position. In this cse we know that the ball falls from a height of 3m, then po = 3m
The position equation is:
p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m
The ball will hit the ground when p(t) = 0m, then we need to solve the equation:
p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m = 0m
for t.
-(1/2)*(9.8m/s^2)*t^2 + 3m = 0m
3m = (1/2)*(9.8m/s^2)*t^2
3m*2 = (9.8m/s^2)*t^2
6m/(9.8m/s^2) = t^2
√(6m/(9.8m/s^2)) = t = 0.78s
The ball needs 0.78 seconds to hit the ground.
The correct answer is (d) the weight of steel ball is much larger than air resistance.
since the density of steel ball is quite higher than that of air. The weight of even a small steel ball will be much larger than the air resistance acting opposite to the motion of the steel ball. Hence in the case of a freely falling steel ball the air resistance can be neglected.
Learn more about air resistance:
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A man walks 10 km north, turns around and walks 5 km south. What displacement did the man walk?
A. 10km S
B. 5km N
C. 15km N
D. 5km S
Answer:
The total displacement is 5 km North (answer B)
Explanation:
Recall that displacement is defined by the vector formula:
Displacement = Final position - initial position
Which corresponds to a vector with its tail at the starting point, and the arrow at the final position. This gives a vector pointing North, and of length 5 km.
This agrees with answer labeled B.
A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the rock after it has fallen halfway to the ground.
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J
look at the diagram below
Which statement is correct about Ray 1?
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the fraction of the legth tof the rod above water
Answer:
[tex]\frac{h_{liquid} }{ h_{body} }[/tex] = 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
[tex]\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }[/tex]
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
[tex]\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }[/tex]
we substitute
5/9 = [tex]\frac{h_{liquid} }{ h_{body} }[/tex]
A fireworks mortar is launched straight upward from a pool deck platform 4 m off the ground at an initial velocity of 61 m/sec. The height of the mortar can be modeled by where h(t) is the height in meters and t is the time in seconds after launch. What is the maximum height
Answer:
22.48m
Explanation:
Given that the pool deck platform is 4 m above the ground.
The initial velocity of mortar in the upward direction, u=21 m/s
At the maximum height, the velocity of the mortar will become zero.
So, the final velocity, v=0
Acceleration due to gravity, g = 9.81 m/[tex]s^2[/tex] in the downward direction.
By using the equation of motion [tex]v^2=u^2+2as\\[/tex]
On putting all the values, we have
[tex]0^2=21^2+2(-9.81)s\\\\s=21^2/(2\times 9.81)[/tex]
s= 22.48 m
Hence, the mortar will reach a maximum height of 22.48m.
1. The Age of the Dinosaurs
Dinosaurs existed about 250 million years ago to 65 million years ago. This era is broken up into three periods known as the Triassic, Jurassic and Cretaceous periods. The Triassic Period lasted for 35 million years from 250-205 million years ago. Planet Earth was a very different place back then. All the continents were united to form one huge land mass known as Pangaea. The Jurassic Period was the second phase. The continents began shifting apart. The time scale for this famous period is from 205 to 138 million years ago. The Cretaceous Period was the last period of the dinosaurs. It spanned a time from 138 million to about 65 million years ago. In this period the continents fully separated. However, Australia and Antarctica were still united.
What type of text structure is this?
Please explain why your answer is correct:
Which property of a solid measures how resistant the material is to deformation?
A. Elasticity
B. Hardness
C. Plasticity
D. resilience
Answer: the answer is a
Explanation:
A 3.50-g bullet has a muzzle velocity of 250 m/s when fired by a rifle with a weight of 25.0 N. (a) Determine the recoil speed of the rifle. m/s (b) If a marksman with a weight of 650 N holds the rifle firmly against his shoulder, determine the recoil speed of the shooter and rifle.
Answer:
(a) 0.343 m/s
(b) 0.012 m/s
Explanation:
(a) From the question above,
MV = mv............................... Equation 1
Where M = mass of the rifle, V = recoiling speed of the rifle, m = mass of the bullet, v = velocity of the bullet.
make V the subject of the equation
V = mv/M........................... Equation 2
Given: m = 3.5 g = 0.0035 kg, v = 250 m/s, M = 25 N = 25/9.8 = 2.55 kg.
Substitute into equation 2
V = (0.0035×250)/2.55
V = 0.343 m/s.
(b) Similarly,
(M'+M)V' = mv....................... Equation 3
Where M' = mass of the marksman, V' = recoiling speed of the shooter and rifle
make V' the subject of the equation
V' = mv/(M'+M)................... Equation 4
Given: m = 3.5 g = 0.0035 kg, v = 250 m/s, M = 25 N = 25/9.8 = 2.55 kg, M' = 650 N = 650/9.8 = 66.33 N
Substitute into equation 4
V' = (0.0035×250)/(66.33+2.55)
V' = 0.8125/68.88
V' = 0.012 m/s
The recoil velocity can be obtained using the principle of conservation of linear momentum.
Using the principle of conservation of linear momentum;
momentum before collision = momentum after collision
Mass of the bullet = 3.50-g or 0.0035 Kg
Mass of the rifle = 2.5 Kg
Where;
M1 = mass of rifle
M2 = mass of bullet
u1 =initial velocity of rife
u2 = initial velocity of the bullet
(2.5 × 0) + (0.0035 × 250) = (0.0035 × 0) + (2.5 × v)
0.875 = 2.5 v
v = 0.35 m/s
For the shooter and the rifle;
(67.5 × 0) + (0.0035 × 250) = (0.0035 × 0) + (67.5 × v)
0.875 = 67.5 × v
v = 0.013 m/s
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Two resistors, A and B, are connected in parallel across of a 6V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6V battery, a voltmeter connected across the resistor A measures a voltage of 4V. Find the resistances of A and B.
Answer:
Resistance of A is [tex]6\ \Omega[/tex] and B is [tex]3\ \Omega[/tex]
Explanation:
The voltage across both the resistances will be the same as they are connected in parallel.
V = Voltage = 6 V
[tex]I_B=2\ \text{A}[/tex]
Resistance is given by
[tex]R_B=\dfrac{V}{I_B}\\\Rightarrow R_B=\dfrac{6}{2}\\\Rightarrow R_B=3\ \Omega[/tex]
[tex]V_B=V_b-V_A\\\Rightarrow V_B=6-4\\\Rightarrow V_B=2\ \text{V}[/tex]
Series connection
[tex]V_A=4\ \text{V}[/tex]
The current is constant in series connection
[tex]I=\dfrac{V_B}{R_B}\\\Rightarrow I=\dfrac{2}{3}\ \text{A}[/tex]
[tex]R_A=\dfrac{V_A}{I}\\\Rightarrow R_A=\dfrac{4}{\dfrac{2}{3}}\\\Rightarrow R_A=6\ \Omega[/tex]
The resistance of A is [tex]6\ \Omega[/tex] and B is [tex]3\ \Omega[/tex].
A child pulls a wagon at a constant velocity of 4.0 m/s for 4.0 minutes along a level sidewalk. The child does this applying a 22 N force to the wagon handle, which is inclined at 35 ° to the sidewalk as shown below. How much work does the child do on the wagon ?
288 k J
8.7 kJ
17.3 kJ
2.3 kJ
Answer:
288kj
Explanation:
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has force constant 815 N/m . The coefficient of kinetic friction between the floor and the block is μk=0.40. The block and spring are released from rest and the block slides along the floor.
Required:
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)
Answer:
v = 0.41 m/s
Explanation:
In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.So, we can write the following general equation, taking the initial and final values of the energies:[tex]\Delta K + \Delta U = W_{ffr} (1)[/tex]
Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.⇒ Kf = 1/2*m*vf² (2)The change in the potential energy, can be written as follows:[tex]\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)[/tex]
where k = force constant = 815 N/m
xf = final displacement of the block = 0.01 m (taking as x=0 the position
for the spring at equilibrium)
x₀ = initial displacement of the block = 0.03 m
Regarding the work done by the force of friction, it can be written as follows:[tex]W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)[/tex]
where μk = coefficient of kinettic friction, Fn = normal force, and Δx =
horizontal displacement.
Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:Fn = Fg= m*g (5)Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:[tex]\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)[/tex]
[tex]\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)[/tex]
Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:[tex]\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)[/tex]
[tex]v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)[/tex]The asteroid Icarus orbits the sun like other planets. Its period is about 410 days. What is its mean distance from the sun
Answer:
Mean distance = 1.61 x 10^8 km
Explanation:
Given the following data;
Orbital period for Icarus, T2 = 410 days
To find the mean distance of Icarus, we would use Kepler's third law of motion.
According to Kepler's third law of planetary motion, the square of any planetary body's orbital period (P) is directly proportional to the cube of its orbit's semi-major axis.
Mathematically, it is given by the formula;
[tex] (\frac {T_{1}}{T_{2}})^2 = (\frac {r_{1}}{r_{2}})^3 [/tex]
Where;
T1 & T2 is the orbital period of a planetary object.
r1 & r2 is the mean distance of a planetary object.
Also, we know that the orbital period for earth, T1 = 365 days
Mean distance of earth = 1.49x10^8 km
Substituting into the equation, we have;
[tex] (\frac {365}{410})^2 = (\frac {1.49x10^{8}}{r_{2}})^3 [/tex]
[tex] (\frac {365}{410}})^2 = (\frac {1.49x10^{8}}{r_{2}})^3 [/tex]
[tex] (0.8902)^2 = (\frac {1.49x10^{8}}{r_{2}})^3 [/tex]
[tex] (0.7925) = (\frac {1.49x10^{8}}{r_{2}})^3 [/tex]
Cross-multiplying, we have;
[tex] (r_{2})^3 = \frac {1.49x10^{8}}{0.7925} [/tex]
Taking the cube root of both sides;
[tex] r_{2} = 1.61 * 10^8 km[/tex]
Need help to get this question right!!
Answer:
16 times as strong
Explanation:
From the question given above, the following assumptions were made:
Initial Force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = ¼r
Final force (F₂) =?
Next, we shall obtain a relationship between the force and the distance apart. This can be obtained as follow:
F = GM₁M₂ / r²
Cross multiply
Fr² = GM₁M₂
If G, M₁ and M₂ are kept constant, then,
F₁r₁² = F₂r₂²
Finally, we determine the new force as follow:
Initial Force (F₁) = F
Initial distance apart (r₁) = r
Final distance apart (r₂) = ¼r
Final force (F₂) =?
Fr² = F₂ × (¼r)²
Fr² = F₂ × r²/16
Fr² = F₂r² / 16
Cross multiply
16Fr² = F₂r²
Divide both side by r²
F₂ = 16Fr² / r²
F₂ = 16F
From the calculations made above, we can see that the new force is 16 times the original force.
Thus, the new force is 16 times stronger.
Which of the following is NOT true about essential body fat? A. The human body would not function normally without essential body fat. B. Essential body fat accounts for about 3% of men's total weight. C. The percentage of essential body fat is the same for both males and females. D. Essential body fat is found in one's organs, bones, and muscles. Please select the best answer from the choices provided. A B C D Mark this and return
Answer:
I believe it is A. The human body would not function normally without essential body fat.
Explanation:
Answer:
Cccccccccccc
Explanation:
(1 point) Unknown resistor in voltage divider Suppose that a power supply is connected across two resistors R1 and R2 that are connected in series. The power supply voltage E, the voltage across the second resistor V2, and the first resistance R1 are all known, but R2 is not known. Find an expression for R2 in terms of E,R1,andV2
Answer:
R₂ = V₂R₁/(E + V₂)
Explanation:
By voltage divider, V₂ = ER₂/(R₁ + R₂)
So, cross-multiplying, we have
V₂(R₁ + R₂) = ER₂
expanding the bracket, we have
V₂R₁ + V₂R₂ = ER₂
collecting like terms, we have
V₂R₁ = ER₂ + V₂R₂
factorizing, we have
V₂R₁ = (E + V₂)R₂
dividing through by (E + V₂), we have
R₂ = V₂R₁/(E + V₂)
Which observation is the best evidence that some colors of visible light are
being absorbed in this photo?
A. The man's head is distorted under the water.
B. The top of the man's head appears disconnected.
C. The shape of the man's arm is clearly seen underwater.
D. The snorkel in the man's mouth appears yellow.
Answer:
D-The snorkel in the mans mouth appears yellow
Explanation:
Answer D is the only example related visible light being absorbed.
An object on the end of a spring with spring constant k moves in simple harmonic motion with amplitude A and frequency f. Which of the following is a possible expression for the kinetic energy of the object as a function of time t?
a. kA^2sin^(2πft)
b. 1/2kA^2cos^2(2πft)
c. 1/2kA sin(2πft)
d. kAcos(2πft)
e. 1/2kAsin(2πft)
The expression for the kinetic energy of the object as a function of time, t is [tex]K.E = \frac{1}{2} k A^2 \ cos^2 \ (2\pi ft)[/tex].
The general wave equation is given as;
[tex]x = A \ cso(\omega t)\\\\x = A \ cos(2 \pi ft)[/tex]
Apply the principle of conservation of energy, the kinetic energy of a particle in such motion is given as;
[tex]K.E = U_x\\\\K.E = \frac{1}{2} kx^2[/tex]
Substitute the value of x into the kinetic energy equation
[tex]K.E = \frac{1}{2} kx^2\\\\K.E = \frac{1}{2} k ( A \ cos (2\pi ft)^2\\\\K.E = \frac{1}{2} k A^2 \ cos^2 \ (2\pi ft)[/tex]
Thus, the expression for the kinetic energy of the object as a function of time, t is [tex]K.E = \frac{1}{2} k A^2 \ cos^2 \ (2\pi ft)[/tex].
Learn more here:https://brainly.com/question/13736293
Once again, move the balloon to the right and let it go. Note how fast the balloon moves. Next, brush the balloon against the entire sweater. Allow all the electrons to transfer. Again, move the balloon all the way to the right, let it go, and note how fast it moves. Is there a difference in how fast the balloon moves when the balloon has more electrons and the sweater has fewer electrons?
Answer:
Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons
Explanation:
From Plato. Hope this helps!
Answer:
Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons
Explanation:
Edmentum
John is conducting an experiment that involves melting ice cubes. Which of the following is most important for John to collect reliable data?
A.The outcome needs to be controlled.
B.An unbiased observer must witness the experiment.
C.Technology needs to be used to determine the results.
D.Only one variable should be tested during the experiment.
Answer:
Hmmmm
Explanation:
Answer:
B
Explanation:
A soccer has been kicked as far as it can get with an initial momentum of 153 kg*m/s and the
ball weights 1.8 kg. What is the velocity of the ball?
Answer:
85 m/s
Explanation:
The formula for momentum is product of weight and velocity
p=mv where m is mass and v is velocity
Given that;
Momentum = 153 kg*m/s
Weight = 1.8 kg
Velocity = ?
p=mv
153 = 1.8 * v
153/1.8 = v
85 m/s
A 845kg dragster (very fast car) accelerates from 2m/s to 30m/s in 0.9s. Determine the average force exerted on the dragster.
Plis I need help
Answer:
Explanation:
Just use the Force formula.
F = M . A
Acceleration Formula
A = V - Vo / Time
So...
F = 845 . (30 - 2 / 0.9)
F = 845 . 20
F = 16900 N
Does the mass of an object make it need more force to move, and to stop?
Answer:
Yea
Explanation:
Let's compare a brick to a small table, does it take more force and strength to push a car or a small table? A car, because it's heavier and has more mass.
Answer:
Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.
Explanation:
☝️
Which hormone is released by the pancreas to reduce blood glucose levels?
Glucagon
Insulin
Glycogen
Answer:
Insulin.
Explanation:
The most important hormone that the pancreas produces is insulin. Insulin is released by the 'beta cells' in the islets of Langerhans in response to food. Its role is to lower glucose levels in the bloodstream and promote the storage of glucose in fat, muscle, liver and other body tissues.
If a student wishes to conduct an experiment to prove the conservation of momentum between two colliding objects, what are the minimum quanitities the student must record in order to complete this experiment and explain why you think they must record those.
Please help with a simple Physics question!
Regions of compression and rarefaction help define ______.
a. electromagnetic waves
b. longitudinal waves, but not transverse waves
c. transverse waves, but not longitudinal waves
d. all mechanical waves
SERIOUS ANSWERS PLEASE, THANK YOU AND GOD BLESS
Compression- a region in a longitudinal (sound) wave where the particles are closest together. Rarefaction- a region in a longitudinal (sound) wave where the particles are furthest apart. Wave motion and particles.
Answer is B.
