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Anon25 [30]
3 years ago
12

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

he sled and cole together is 100 kg. how far does the sled go before stopping? use the formula: a= (vf2-vi2)/2d regarding the relation among acceleration a, final velocity vf, initial velocity vi and the traveled distance of an object
d.
Physics
1 answer:
stepan [7]3 years ago
3 0
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
v_f=0 is the final speed of the sled
v_i=5 m/s is the initial speed
d is the distance covered

By rearranging the equation, we find d:
d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m
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A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric
Alex73 [517]

Answer:

(a) -6.76\times 10^{12}\ N/C

(b) -1.352\times 10^{13}\ N/C

(c) -7.2\times 10^{11}\ N/C

Explanation:

(a)

Given:

Charge on sphere (Q) = -20\ \mu C=-20\times 10^{-6}\ C

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = 9\times 10^{9}\ Nm^2/C^2

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

E=(\frac{kQ}{R^3})r

Plug in the given values and solve for 'E'. This gives,

E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)

(b)

Given:

Charge on sphere (Q) = -20\ \mu C=-20\times 10^{-6}\ C

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

E=(\frac{kQ}{R^3})r

Plug in the given values and solve for 'E'. This gives,

E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)

(c)

Given:

Charge on sphere (Q) = -20\ \mu C=-20\times 10^{-6}\ C

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

E=\dfrac{kQ}{r^2}

Plug in the given values and solve for 'E'. This gives,

E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)

8 0
4 years ago
Determine the apple's kinetic energy, KE, the gravitational potential energy of the system, gravity, and the total mechanical en
guajiro [1.7K]

Answer:

7.45

Explanation:

At 3.0 m PE = m*g*3.0, KE = m*g*(4.0 - 3.0)

At 2.0 m PE = m*g*2.0, KE = m*g*(4.0 - 2.0)

At 3.0 m PE = m*g*1.0, KE = m*g*(4.0 - 1.0)

At 0 m PE = 0, KE = m*g*h = total energy = 7.45 J

6 0
3 years ago
A rope pulls a 20 kg box up. If the tension in the rope is 275 N, what is the acceleration of the box?
Nana76 [90]

Answer:

<h3>The answer is 13.75 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{275}{20}  =  \frac{55}{4}  \\

We have the final answer as

<h3>13.75 m/s²</h3>

Hope this helps you

4 0
3 years ago
When sighting a firearm using an open sight, in what direction should the rear sight be moved if you want the shot to travel to
sattari [20]

Explanation:

Test to see where the bullets gather on your target after shooting a number of shots. Change the targets of your weapon if the bullets are not in the target's middle. Try to shift the rear view in the same direction with an open sight that you would like to push the shots on the target.

After the rifle is sighted-in, one must practice firing under various lighting and weather conditions, and at different distances to be able to become accurate and proficient shooter.

3 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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