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3 years ago

A train moves from rest to a speed of 25 m/s in 30.0 seconds . What is the acceleration

1 answer:

7 0

When a car moves from rest then the initial velocity,
$u = 0m {s}^{ - 1}$

The final velocity,

$v = 25m {s}^{ - 1}$
and time,
$t = 30s$
Using the equation,
$v = u + at$
We have,

$25 = 0 + a(30)$
$25 = 30a$
$\frac{25}{30} = a$
$\therefore \: a = \frac{5}{6} m {s}^{ - 2}$

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While driving a 2150 kg car, carl steps on the gas pedal, accelerating at 4.0 m/s2. If the coefficient of friction between his c

Mrac [35]

Answer:

d. 3332.5 [N]

Explanation:

To solve this problem we will use newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.

Here we have two forces, the force that pushes the car to move forward and the friction force.

The friction force is equal to the product of the normal force by the coefficient of friction.

f = N * μ

f = (m*g) * μ

where:

N = weight of the car = 2150*9.81 = 21091.5 [N]

μ = 0.25

f = (21091.5) * 0.25

f = 5273 [N]

Now as the car is moving forward, the car wheels move clockwise. The friction force between the wheels of the car and the pavement must be counterclockwise, i.e. counterclockwise. Therefore the direction of this force is forward. This way we have:

F + f = m*a

F + 5273 = 2150*4

F = 8600 - 5273

F = 3327 [N]

Therefore the answer is d.

6 0

3 years ago

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v

Brut [27]

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

$l = 2a$

The distance of the center of the square from each of the corners is

$r = \frac{2a}{2}= a$

The potential electric at the center due to each cornet charge is

$V_1 = \frac{kQ_1}{r_1}$

$V_2 = \frac{kQ_2}{r_2}$

$V_3 = \frac{kQ_3}{r_3}$

$V_4 = \frac{kQ_4}{r_4}$

The total electric potential at the center of the given square is

$V = V_1+V_2+V_3+V_4$

$V = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}$

Al the charges are equal, and the distance are equal to a, then

$V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}$

$V = \frac{4kQ}{a}$

Therefore the correct option is E.

3 0

3 years ago

Vector B~ has x, y, and z components of 8.7,

blagie [28]

Answer:

The magnitude of B is $10.95\ units$

Explanation:

we know that

The magnitude of Vector B is

$B=\sqrt{x^{2}+y^{2}+z^{2}}$

where

x,y and z are the components of vector B

we have

$x=8.7\ units, y=1.4\ units,z=6,5\ units$

substitute

$B=\sqrt{8.7^{2}+1.4^{2}+6.5^{2}}$

$B=\sqrt{119.9}$

$B=10.95\ units$

3 0

3 years ago

A golfer badly misjudges a putt, sending the ball only one-quarter of the distance to the hole. The original putt gave the ball

IRISSAK [1]

Answer:

v = 2 v₀

Explanation:

For this exercise we will use the relationship between work and the change in kinetic energy

W = ΔK

Where the work is of the friction force, which is always opposed to the movement whereby the angle is 180º

W = - fr x

The distance for the first case is

x = 1/4 L

We substitute

-fr ¼ L = ½ m v₀²

fr L = 2 m v₀²

In the second case, the new speed takes the ball the entire distance

x = d

-fr L = ½ m v²

We equal the two equations

2 m v₀² = ½ m v²

v² = 4 v₀²

v = 2 v₀

3 0

3 years ago

A boat with an anchor on board floats in a swimming pool that is somewhat wider than the boat. Does the pool water level move up

pentagon [3]

Answer:

a) moves down

b) moves down

c) level remains same

Explanation:

Given that the anchor is initially on the floating boat.

In this condition initially the the volume of water $(V_w_i)$ displaced is to balance its weight.

Now,

$W_a=W_w$

$V_a.\rho_a.g=V_w_i.\rho_w.g$

$\frac{\rho_a}{\rho_w} =\frac{V_w_i}{V_a}$

We've, the density of steel $= 7850\ kg.^{-3}$ and the density of water $= 1000\ kg.^{-3}$

$\therefore V_w_i=7.85\times V_a$

When the anchor is dropped into water:

The volume of water displaced be $(V_w_f)$ which will be equal to the volume of anchor since it is immersed into it.

$V_a=V_w_f$

$\therefore V_w_i>V_w_f$ ...................(1)

So the level of water falls when the anchor is dropped into water.

Now, when the anchor is thrown on the ground the water has now less weight to balance so the water level falls down.

When the cork on the from the boat is dropped into the water and it still floats then it must displace same amount of water, hence there should be no change in the water level.

6 0

3 years ago