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3 years ago

Which of the following is an example of Newton's second law of motion?

Physics

2 answers:

laiz [17]3 years ago

5 0

Answer:

Explanation:

Newton’s Second Law of Motion

Newton’s Second Law of Motion states that ‘when an object is acted on by an outside force, the mass of the object equals the strength of the force times the resulting acceleration’.

This can be demonstrated dropping a rock or and tissue at the same time from a ladder. They fall at an equal rate—their acceleration is constant due to the force of gravity acting on them.

The rock's impact will be a much greater force when it hits the ground, because of its greater mass. If you drop the two objects into a dish of water, you can see how different the force of impact for each object was, based on the splash made in the water by each one.

uysha [10]3 years ago

4 0

Answer: B

Explanation:

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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to

aksik [14]

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

$\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y = \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} } =2\ m/s$

Also:

$\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x = u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s$

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

$v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s$

4 0

3 years ago

Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago

The use of force to move an object is <br> .

Ne4ueva [31]

The use of force to move an object is called work. This only applies if the object moves.

4 0

3 years ago

a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r

dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

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4 0

2 years ago

Name the property illustrated be each statement. If t - 13 = 52, then 52 = t - 13

Natali5045456 [20]

The property illustrated to each statement is commutative because it shows that for some value of t both statements are equal.

3 0

3 years ago

Read 2 more answers