Do 2. A cyclist starts from rest and accelerates along a straight path to a speed of
12.15 m/s in a time of 4.5 seconds.
What is the cyclist's acceleration to the nearest tenth?
O
54.7 m/s2
o
24.3 m/s?
2.7 m/s2
De
3.4 m/s?

Answer:

50

Explanation:

Answer:A

Explanation:

Absolute method

Answer:

absolute method

Explanation:

trust

Answer:

C. Wanda's waist circumference indicates that she is not at risk, but her BMI indicates that she underweight. She should develop a plan to move her BMI into the normal range.

Explanation:

edge 2020 Brainliest would be appreciated

Answer:

Wanda's waist circumference indicates that she is not at risk, but her BMI indicates that she underweight. She should develop a plan to move her BMI into the normal range.

Explanation:

I just did the Cumulative Exam Review

Answer:

if the force increases while the mass of the ball remains the same, the acceleration increases.

Explanation:

According to Newton's second law of motion, the acceleration of an object produced by a net force is directly proportional to the magnitude of the net force Fn and inversely proportional to the mass of the object m.

[tex]\displaystyle a=\frac{Fn}{m}[/tex]

Suppose the net force doubles to 2Fn. The new acceleration will be:

[tex]\displaystyle a'=\frac{2Fn}{m}=2\frac{Fn}{m}=2a[/tex]

The acceleration doubles too. Thus, if the force increases while the mass of the ball remains the same, the acceleration increases.

Answer:

am pretty sure the answer is d

Answer:

90 km

Explanation:

45 km per hr

2 hrs

45 x 2 = 90

90 km/2 hrs

Hope this helps!

Answer:

The pressure in the tunnel is 97.45 kPa

Explanation:

Given;

atmospheric pressure, P₁ = 101.3 kPa

air velocity in the tunnel, v₂ = 80 m/s

density of air at 20°C, ρ = 1.204 kg/m³

pressure in the tunnel, P₂ = ?

Apply Bernoulli's equation;

P₁ + ¹/₂ρv₁² + ρgz₁ = P₂ + ¹/₂ρv₂² + ρgz₂

v₁ = 0

z₁ = z₂ = 0

P₁ = P₂ + ¹/₂ρv₂²

P₂ = P₁ - ¹/₂ρv₂²

P₂ = 101.3 kPa - (0.5 x 1.204 x 80²)

P₂ = 101.3 kPa - 3.8528 kPa

P₂ = 97.45 kPa

Therefore, the pressure in the tunnel is 97.45 kPa

Answer:

124m/s

Explanation:

v=u+at

v=119+(0.5×10)

v=119+5

v=124m/s

Earth's rotation is crucial to understanding the Coriolis effect. The Equator is where Earth rotates more quickly than the poles. There at Equator, Earth is larger.

Equatorial areas therefore move at a speed of around 1,600 km/h (1,000 mph) in a 24-hour period to complete a revolution. Earth rotates slowly towards to the poles, at 0.00008 km/h (0.00005 mi/h).

Low-pressure systems called cyclones draw air towards their "eye" or central region. Fluids representing high-pressure systems in the Northern Hemisphere pass low-pressure regions to the right.

Air masses are deflected when they're drawn inside cyclones from across all sides, giving the appearance that a hurricane is rotating clockwise. Trade winds refer to these air masses' recurrent circulation patterns.

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Answer:

The answer is 4, the west.

Answer:

The west

Explanation:

hopes this helps you and your questions

Answer:

60 kg

Explanation:

An astronaut weighs 588 N on the earth

He also weighs 98 N on the moon

Therefore the the mass of the astronaut on the moon can be calculated as follows

= 588/98

= 6 × 10

= 60 kg

Hence the mass of the astronaut on the moon is 60 kg

The mass of the astronaut on the moon will be equal to 60 kg whereas the gravitational acceleration on the moon is 1.63 [tex]\rm m/s^2[/tex].

Given information:

An astronaut weighing 588 N on the earth notices that he weighs only 98 N on the moon.

The mass of any object in the whole universe doesn't change but the gravitational force or weight can change. This is because the gravitational acceleration can change.

The mass of the object in the earth or anywhere will be,

[tex]w=mg\\588=m\times 9.8\\m=60\rm \; kg[/tex]

So, the astronaut has 60 kg mass which is constant.

The gravitational acceleration on the moon will be,

[tex]98=60a\\a=1.63\rm\; m/s^2[/tex]

Therefore, the mass of the astronaut on the moon will be equal to 60 kg whereas the gravitational acceleration on the moon is 1.63 [tex]\rm m/s^2[/tex].

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Answer: $78.50 or C.

Explanation:

$92.35 x 15% or .15 is $78.50

Answer:

F = 3600 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force [N]

m = mass = 2000 [kg]

a = acceleration = 1.8 [m/s^2]

Now replacing:

F = 2000*1.8

F = 3600 [N]

Answer:

Explanation:

Given

initial velocity u = 0m/s

final velocity v = 18m/s

To get the time it takes the bird to make this trip, we will use the equation of motion

v = u+gt

g is the acceleration due to gravity = 9.81m/s²

t is the time

Substitute into the formula:

[tex]18 = 0+(9.81)t\\18 = 9.81t\\9.81t = 18\\t = \frac{18}{9.81}\\ t = 1.84secs[/tex]

Hence it took the bird 1.84seconds to make this trip

Answer:

t=0.14s

Explanation:

Answer:

The father will walk 6 more meters and the son will walk 12 meters

Explanation:

Uniform Speed and Acceleration

This is a problem where two objects have different types of motion. The father walks at a constant speed and later, his son starts a constant acceleration motion in pursuit of him.

Let's start with the father, whose speed is v=1 m/s during t=6 seconds. He travels a distance:

[tex]x=vt=1*6=6\ m[/tex]

Now the son, starting from rest (vo=0) accelerates at a=2/3 m/s^2. His speed will increase and eventually, he will catch up with his father. Let's suppose it happens at a time t1.

The distance traveled by the son is given by:

[tex]\displaystyle xs=v_o.t_1+\frac{a.t_1^2}{2}[/tex]

Since vo=0:

[tex]\displaystyle xs=\frac{a.t_1^2}{2}[/tex]

The father will continue with constant speed and travels a distance of:

[tex]xf=v.t_1[/tex]

For them to catch up, the distance of the son must be 6 m more than the distance of the father, because of the leading distance he has already taken. Thus:

[tex]xs=xf+6[/tex]

Substituting the equations of each man:

[tex]\displaystyle \frac{a.t_1^2}{2}=v.t_1+6[/tex]

We know a=2/3, v=1:

[tex]\displaystyle \frac{2}{3}\cdot\frac{t_1^2}{2}=t_1+6[/tex]

Simplifying:

[tex]\displaystyle \frac{t_1^2}{3}=t_1+6[/tex]

Multiply by 3:

[tex]t_1^2=3t_1+18[/tex]

Rearranging:

[tex]t_1^2-3t_1-18=0[/tex]

Factoring:

[tex](t_1-6)(t_1+3)=0[/tex]

Solving:

[tex]t_1=6 , t_1=-3[/tex]

Since time cannot be negative, the only valid solution is

[tex]t_1=6\ s[/tex]

The distance traveled by the son in 6 seconds is:

[tex]\displaystyle xs=\frac{2/3\cdot 6^2}{2}[/tex]

[tex]xs=12\ m[/tex]

Note the father will travel

[tex]xf=1*6=6\ m[/tex]

This 6 m plus the 6 m he was ahead of the son, make them meet while walking at 6 seconds.

Answer: The father will walk 6 more meters and the son will walk 12 meters

Analysing the question:

We are given:

mass of object (m) = 0.03 kg

height of object (h) = 2.94 m

acceleration due to gravity (a) = 10 m/s²

initial velocity (u) = 0 m/s [the object fell, we are NOT given any initial velocity]

final velocity (v) = v m/s

Solving for Final velocity:

from the third equation of motion:

v² - u² = 2ah

replacing the variables

(v)² - (0)² = 2(10)(2.94)

v² = 58.8

v = 7.67 m/s

The distance travelled in total is 14 m.  But the resultant displacement is the square root of the sum of squares of each distances covered in different directions, that is 10 m.

Displacement is a physical quantity that, that measures how far we travelled from a mean position. The change in position is dependent on the magnitude and direction as well. Hence, displacement is a vector quantity.

The distance covered by an object is total length or sum magnitudes of the length it covered. It is a scalar quantity. Here, the distance covered is 6 + 8 = 14 m.

The resultant displacement here, is calculated as follows:

ds = √(6² + 8²) = √100 = 10 m

Therefore, the displacement of us will be 10 m.

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Answer:

The potential difference is   [tex]\Delta V  =  44.40 \  V [/tex]

Explanation:

From the question we are told that

   The diameter of the droplet of oil is [tex]d =  0.71 \mu m  =  0.71 *10^{-6} \ m[/tex]

    The density of the oil is  [tex]\rho =  860 kg/m^3[/tex]

    The distance of separation of the capacitor plate is  [tex]l  = 4.5 \ mm =  0.0045 \  m[/tex]

Generally the radius of the droplet is mathematically represented as

       [tex]r = \frac{d}{2}[/tex]

=>      [tex]r = \frac{0.71 *10^{-6} }{2}[/tex]

=>      [tex]r = 3.55 *0^{-7} \  m[/tex]

Generally the mass of  the oil droplet is mathematically represented as

       [tex]m  =  \rho  *  V[/tex]

Here  V  is the volume of the oil droplet which is mathematically represented as

      [tex]V  =  \frac{4}{3} * 3.142 * (3.55 *0^{-7} )^3[/tex]

       [tex]V  =  1.874 *10^{-19} \  m^3[/tex]

So

       [tex]m  =  860  *  1.874 *10^{-19} [/tex]

=>    [tex]m  =  1.611 *10^{-16} \  kg [/tex]

Generally the electric force acting on the droplet is mathematically represented as  

       [tex]F  =  E  *  q[/tex]

Here q is the charge on an electron with value  [tex]q =  1.60*10^{-19}\ C[/tex]

This force is equivalent to the weight of the droplet which is mathematically represented as

        [tex]W =  mg[/tex]

So

       [tex]E *  q =  m *  g[/tex]

Here E is the electric field which is mathematically represented as

    [tex]E =  \frac{\Delta V}{l}[/tex]

      [tex]\frac{\Delta V}{l}  *  q =  m *  g[/tex]

=>    [tex]\Delta V  =  \frac{m *  g  *  l }{q}[/tex]

=>    [tex]\Delta V  =  \frac{1.611 *10^{-16}  *  9.8   *  0.0045 }{1.60*10^{-19}}[/tex]

=>    [tex]\Delta V  =  44.40 \  V [/tex]

Acceleration is the change in the velocity of the object with time. Acceleration is the same for both objects so both objects will reach the ground at the same time.

Given that both the object dropped at the same time. So the gravitational acceleration of both the objects will be the same which is nearly 10 m/s.

The force at object 1 is given below.

[tex]F_1 = m_1\times g = m_1\times a_1[/tex]

Where m1 is the mass of object 1 and a1 is the acceleration of object 1.

[tex]F _1 = 1\times 10 = 10 \;\rm N[/tex]

The acceleration of object 1 is calculated as below.

[tex]a_1 = \dfrac{F_1}{m_1}[/tex]

[tex]a_1=\dfrac{10}{1} = 10 \;\rm m/s^2[/tex]

The force at object 2 is given below.

[tex]F_2 = m_2\times g = m_2\times a_2[/tex]

Where m2 is the mass of object 2 and a2 is the acceleration of object 2.

[tex]F_2 = 2\times 10 =20 \;\rm N[/tex]

The acceleration of object 2 is calculated as below.

[tex]a_2 = \dfrac{F_2}{m_2}[/tex]

[tex]a_2=\dfrac{20}{2} = 10 \;\rm m/s^2[/tex]

Hence we can conclude that the acceleration of both the objects is the same at every point so they will reach the ground at the same time.

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Answer:

Electromagnetic waves transfer energy through vibrations of electric and in magnetic fields. For sound waves the energy is transferred through the vibration of air particles or particles of a solid. Waves also carry energy, the amount of energy they carry depends on their frequency and their amplitude. The more energy and the higher frequency amplitude, the more energy waves are ^‿^

Answer:

i think radian degree thank you

Explanation:

cuz right the degree angle i responsible more

Answer:

Its radian degree

Explanation:

Because the length of an arc of a unit circle is equal to measurement in radians of angle.

So it is the biggest unit of plane angle

Answer:

4

Explanation:

From the question given above, the following data were obtained from:

Length of an axe = 20 cm

Width = 5 cm

Mechanical advantage (MA) =.?

The mechanica advantage is simply defined as the ratio of the length to the width of the axe. Mathematically, it is expressed as:

Mechanical advantage = length /width

With the above formula, we can obtain the mechanical advantage of the axe as follow:

Length of an axe = 20 cm

Width = 5 cm

Mechanical advantage (MA) =.?

MA = 20/5

MA = 4

Therefore, the mechanical advantage (MA) of the axe is 4

To find:-

Answer:-

Gauss's law in electrostatics states that the electric flux through a closed surface is equal to 1/ε0 times the charge enclosed by that surface, that is,

[tex]\implies \displaystyle \phi_E = \oint \vec{E} .\vec{ds} \\[/tex]

Also ,

[tex]\implies\displaystyle \oint \vec{E}.\vec{ds} =\dfrac{q}{\epsilon_0} \\[/tex]

Linear density of charge:-

It ise defined as charge per unit length. Mathematically,

[tex]\implies\displaystyle \lambda =\dfrac{q}{\ell} \\[/tex]

where ,

For figure see attachment.

Now let us assume that the linear density of charge is [tex]\lambda[/tex] . There is a point P near the wire at a distance of " r " from it , where we are interested in calculating the electrical field. Imagine a coaxial Gaussian cylinder of length [tex]\ell[/tex] , and radius " r " such that the point P lines on its surface.

Calculating Electrical flux through the cylinder:-

[tex]\implies\displaystyle \phi_E = 0 + 0 + \oint E.ds \\[/tex]

[tex]\implies\displaystyle\phi_E = \oint E.ds \ \cos0^o \\[/tex]

[tex]\implies\displaystyle \phi_E = \oint Eds \\[/tex]

[tex]\implies\displaystyle \phi_E = E\oint ds \\[/tex]

[tex]\implies\displaystyle \phi_E = E (2\pi r\ell) \ \ dots (1) \\[/tex]

From Gauss's law:-

[tex]\implies\displaystyle \phi_E =\dfrac{q}{\epsilon_0} \\[/tex]

[tex]\implies\displaystyle \phi_E = \dfrac{\lambda \ell}{\epsilon_0}\ \dots (2) \\[/tex]

Therefore , from equation (1) and (2) , we have;

[tex]\implies\displaystyle E \times 2\pi r \ell =\dfrac{\lambda\ell}{\epsilon_0}\\[/tex]

[tex]\implies\displaystyle \underline{\underline{\green{ E =\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda}{r}}}}\\[/tex]

And we are done!

Answer:

The diameter of a hydrogen atom based on scientific notation is 1.06 x 10^-10 m

The diameter of a hydrogen atom is 0.000000000106 m. This is number can best be expressed in scientific notation as 1.06 x 10⁻¹⁰ m. Hence, option (C) is correct.

Scientific notation is a means to express values that are either too big or too little to be conveniently stated in decimal form (typically would result in a long string of digits). It is also known as standard form in the UK and scientific form, standard index form, and standard form.

Scientists, mathematicians, and engineers frequently utilize this base ten notation because it can make some mathematical operations simpler. It is typically referred to as "SCI" display mode on scientific calculators.

As there has 9 zeros after the decimal, this number can best be expressed in scientific notation as 1.06 x 10⁻¹⁰ m.

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Answer:

The answer of this question is windmill

Answer:

1.5 hr

16.7

Explanation:

Zero apparent weight means there's no normal force.

Sum the forces in the centripetal direction.

∑F = ma

mg = mv²/r

v = √(gr)

v = √(7.4×10⁶ m × 10 m/s²)

v = 8602 m/s

The circumference of the equator is:

C = 2πr

C = 2π (7.4×10⁶ m)

C = 4.65×10⁷ m

So the period is:

T = C / v

T = (4.65×10⁷ m) / (8602 m/s)

T = 5405 s

T = 1.5 hr

The initial speed is:

v = C / T

v = (4.65×10⁷ m) / (25 h × 3600 s/h)

v = 517 m/s

The speed increases by a factor of:

8602 m/s / 517 m/s

16.7

The new period of the planet at the given radius is 1.5 h.

The speed of the planet increased by a factor of 16.7.

The given parameters;

The speed of the planet at the given radius is calculated as follows;

[tex]mg = \frac{mv^2}{r} \\\\gr = v^2\\\\v= \sqrt{rg} \\\\v = \sqrt{7.4 \times 10^6 \times 10} \\\\v = 8602.3 \ m/s[/tex]

The new period of the planet is calculated as follows;

[tex]T = \frac{2\pi R}{v} \\\\T = \frac{2 \pi \times 7.4 \times 10^6}{8602.3} \\\\T =5,405.72 s \\\\T = \frac{5405.72}{3600} = 1.5 \ hr[/tex]

The initial speed of the planet is calculated as follows;

[tex]v_i = \frac{2\pi R}{T} \\\\v_i = \frac{2\pi \times 7.4\times 10^6}{25 \times 3600} \\\\v_i =516.68 \ m/s[/tex]

The increase in the speed of the planet is calculated as;

[tex]= \frac{v}{v_i} \\\\= \frac{8602.3}{516.68} \\\\= 16.7[/tex]

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