Ask question

Ask question

All categories

3 years ago

12

18. A person pulls horizontally with a force of 64 N on a 14-kg box. There is a force of friction between the box and the floor

of 36 N. Find the acceleration of the box in m/s2.

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

since force=massxacceleration
therefore acceleration=force/mass
the resultant force=the acting force on the object-the frictional force
the resultant force=64-36=28N
therefore the acceleration=28/14=2m/sec^2

You might be interested in

Riders on a ferris wheel move in a circle with a speed of 4.0 m/s. As they go around, they

Vanyuwa [196]

Answer: D = 16m

Explanation: given values: a = 2 m/s2, v = 4 m/s
In this case we have to determine the diameter of the Ferris wheel.
Diameter of circle is given as: D = 2.r.
First we have to find radius of wheel. The best way to find that is using the centripetal acceleration equation: a = v2/r
Plug in values in above equation to find radius: 2 m/s2 = (4 m/s)2/r 2 m/s2 = (16 m2/s2)/r r = (16 m2/s2)/2 m/s2
r = 8.0m
Diameter of Ferris wheel is:
D = 2.r.
D = 2.8m
D = 16m

6 0

3 years ago

frosja888 [35]

B, refraction I believe.

5 0

3 years ago

Identify two structural characteristics specific to muscle tissue.

Semmy [17]

Answer:

Explanation:

Muscle cells are excitable; they respond to a stimulus.meaning they can shorten and generate a pulling force. When attached between two movable objects, such as two bones, contraction of the muscles cause the bones to move.It contains protein fibers which contract to make the cell shorter.

7 0

3 years ago

Read 2 more answers

The acceleration of gravity on the moon is about 1.6 m/sec2. An experiment to test gravity compares the time it takes objects to

wlad13 [49]

I really think that the answer A

3 0

3 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago