Help if u understand for a b c

Answer:

Explanation:

F=ma

F in this case is the gravity acting on the 2kg object. Acceleration of gravity is 9.8 m/s^2. SF=ma, so

F = 2kg*(9.8 m/s^2) = 19.6 N

Now use this force to determine the mass of the object on the table:

F=ma

19.6 N (1N=kg*m/s^2) = m*(1.8 m/s^2)

m = 10.89 kg

Let volume of empty boat be = 100% = 1V

and mass of boat be M

In water 10%, 0.1V of the volume is submerged.

Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V

M leads to 0.1V boat submerging

boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

120kg = 0.6M

M = (120/0.6)kg

M = 200kg.

The mass of the boat is 200kg.

Answer:

Work equals to force times distance

Explanation:

the equation would be work=force*distance

Answer:

5

Explanation:

Formula for Mass (for this question)

[tex]mass=\frac{Force}{Acceleration}[/tex]

mass= 1000N/200[tex]m^{2}[/tex]=5

(pls do correct me if I have any mistakes it would rlly help others!)

Answer:

NO Y

Explanation:

MMMMMMMMM

The unknown variables in the model equation for ball's motion include vertical distance traveled by the ball which is 31.86 m and the final velocity of the ball which is 25 m/s.

The given parameter;

The equations that model the motion of the ball is given as follows;

[tex]h = v_0_y + \frac{1}{2} gt^2[/tex]

[tex]v_f = v_0_y + gt[/tex]

All the unknown variables in the equations include;

The vertical distance traveled by the ball is calculated as follows;

[tex]h = \frac{1}{2} gt^2\\\\h = 0.5 \times 9.8 \times 2.55^2\\\\h = 31.86 \ m[/tex]

The final velocity of the ball is calculated as follows;

[tex]v_f = 0 + 9.8(2.55)\\\\v_f = 25 \ m/s[/tex]

Thus, the unknown variables in the model equation for ball's motion include vertical distance traveled by the ball which is 31.86 m and the final velocity of the ball which is 25 m/s.

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Answer:

It isn't A.

Explanation:

I got it wrong with the answer A.

it is harder to swim in water with a current (like the ocean) than in contained water (like a pool) because Still water is an immovable object, but currents have their own momentum. The correct option is D.

Newton's first law of motion, also known as the law of inertia, states that "an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an unbalanced force."

This means that an object will continue to move in a straight line at a constant speed unless an unbalanced force acts upon it. In addition, if an object is at rest, it will remain at rest unless acted upon by an unbalanced force.

The law of inertia can be applied to many situations. For example, if a book is sitting on a table, it will remain at rest until a force is applied to it. Similarly, if a ball is rolling on a flat surface, it will continue to roll at a constant speed in a straight line unless a force, such as friction, slows it down or changes its direction.

This law is fundamental to understanding the behavior of objects in motion, and it forms the basis for the other two laws of motion formulated by Sir Isaac Newton.

Here in the Question,

Swimming in a pool versus swimming in the ocean, the difference lies in the presence of a current in the ocean.

The current in the ocean has its own momentum, and when a swimmer enters the water, they are faced with an additional force that they must overcome to move forward. This means that the swimmer must apply more force and use more energy to overcome the force of the current and move forward in the water.

In still water, on the other hand, there is no external force acting on the swimmer, so they only have to overcome the resistance of the water itself to move forward.

Therefore, The correct answer is D i.e Still water is an immovable object, but currents have their own momentum.

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Explanation:

v=?, u=0, a=?, S=22m.

Using the formula, S=ut+½at²

22={0×5}+(½.a.5²)

22=½.a.5²

a=44/25 = 1.76m/s².

Therefore, net force = work done = ma = 48×1.76 = 84.48N.

therefore, power = work done/time = 84.48/5 = 16.896W.

hope this helps you.

Answer:

Gibbs, free energy

Explanation:

[tex]\\ \sf\longmapsto Q=mc\Delta T[/tex]

[tex]\\ \sf\longmapsto c=\dfrac{Q}{m\Delta T}[/tex]

[tex]\\ \sf\longmapsto c=\dfrac{1320}{5.61(274)}[/tex]

[tex]\\ \sf\longmapsto c=\dfrac{1320}{1537.14}[/tex]

[tex]\\ \sf\longmapsto c=0.85J/kgK[/tex]

4.44J (creds to commenter on the post, just making it bigger)

Answer:

Availability of charging

Explanation:

Assuming the car is not a hybrid and your traveling a very long distance far from civilization, you wouldn't be able to charge the car if it runs out of power as no outlets would be around.

Answer:

An object must be touching another.

Answer:

W = 90 J

Explanation:

Data:

Use the formula:

Replace and solve:

The work done was 90 Joules.

Greetings.

Hi there!

We can begin by solving for the time it takes the projectile to reach the ground.

Use the equation:

[tex]t = \sqrt{\frac{2h}{g}}[/tex]

Plug in the given height:

[tex]t = \sqrt{\frac{2(357)}{9.8}} = 8.536 s[/tex]

Use the equation:

d = vt, to solve for the horizontal distance:

d = (54)(8.536) = 460.944 m

Answer:

Plato, Aristotle developed it further and used for 1400 years till Copernicus.

Explanation:

The coefficient of kinetic friction between the chair and the floor is 0.442.

The given parameters;

The normal force on the chair is calculated as follows;

Fₙ = mg

Fₙ = 30 x 9.8 = 294 N

The frictional force between the chair and the floor is calculated as follows;

Fk = μFₙ

The net force on the chair is calculated as follows;

[tex]\Sigma F_x = 0\\\\F- \mu_k F_n = 0\\\\130 - \mu_k (294) = 0\\\\\mu_k = \frac{130}{294} \\\\\mu_k = 0.442[/tex]

Thus, the coefficient of kinetic friction between the chair and the floor is 0.442.

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Answer:

A

Explanation:

Low energy photons (such as radio photons) behave more like waves, while higher energy photons (such as X-rays) behave more like particles.

Answer:

B. It has a short amplitude.

:)

If Kevin with mass 45kg pushes Kisha with a force of 54N, the magnitude of Kevin's acceleration is 1.2 m/s²

Kisha's mass = 60 kg

Kevin's mass = 45 kg

Kevin pushes off Kisha with a force of 54 N

Since it was Kevin that pushed Kisha, and the required force is 54N, we can look for Kevin's acceleration using the equation below

F  =  ma

The considered mass is Kevin's mass

54   =  45a

a   =   54/45

a   =    1.2 m/s²

The magnitude of Kevin's acceleration is 1.2 m/s²

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Answer:

This is Newton’s third law (forces due to friction). There would be 15N being pushed back at Jane.

Explanation:

Newton’s third law states that for every action there is an equal and opposite reaction. So if Jane pushed with a force of 15N, 15N would be pushed back at her. Hope this helps! :)

Addition of vectors:

vector u

has a magnitude of 20 and a direction of 0º with respect to the horizontal, vector v has a magnitude of 40 and a direction of 60º with respect to the horizontal.

a) Find the magnitude and direction of the resultant to the nearest whole

number.

Vector Sum:

The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

Answer and Explanation:

Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

of magnitude 40 and direction 60∘ to the horizontal axis, the vector is →v=^i40cos60∘+^j40sin60∘.

The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

|→w|∠θ=√(40)2+(20√3)2∠tan−1(20√340)≈52.9∠40.9∘

The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

The magnitude of the resultant vector would be 60 units, and its direction will be 0°.

To find the magnitude and direction of the resultant vector when adding vectors u and v, we can use vector addition formulas:

Magnitude of the Resultant Vector (R):

R = √(u^2 + v^2 + 2uv * cosθ)

where:

u = magnitude of vector u (20)

v = magnitude of vector v (40)

θ = angle between vectors u and v (60°)

Plugging in the values:

R = √(20^2 + 40^2 + 2 * 20 * 40 * cos(60°))

≈ √(400 + 1600 + 1600)

≈ √3600

= 60

Direction of the Resultant Vector (θR):

Use the Law of Cosines to find the angle θR between the resultant vector R and vector u:

cosθR = (v^2 + R^2 - u^2) / (2 * v * R)

Plugging in the values:

cosθR = (40^2 + 60^2 - 20^2) / (2 * 40 * 60)

= (1600 + 3600 - 400) / 4800

= 1

Since cos(0°) = 1, we have θR = 0°.

So, the magnitude of the resultant vector is 60 units, and its direction is 0°.

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Answer:

So if the half-life is two days, four half-lives is 8 days. Suppose a sample has a count rate of 3,200 Becquerel (Bq) at the start, then its count rate after 8 days would be 1/16th of 3,200 Bq = 200 Bq.

Answer:

Explanation:

using Charles law; since the pressure is constant

V1/T1 = V2/T2

5/300 = V2/350

cross multiply

V2 X 300 = 5 X 350

V2 = 1750/300

V2 = 5.833L

Answer:

about 16.5 by-

wait DO UR OWN MATH lol

Explanation:

Answer:

p>Whether an object will float or sink is dependent on its density, and on the density of the liquid it is placed in. In the case of water, an object with a density less than 1 g/cm3 will float. The closer its density is to 1 g/cm3, the more of it will sit below the water level.