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hammer [34]
3 years ago
10

How does the potential energy used by the counterweight as it drops relate to the kinetic energy of the projectile as it is laun

ched?
Physics
1 answer:
horsena [70]3 years ago
6 0
When the counter weight is released, it plummets towards the ground causing the projectile on the other end to rotate. As the counterweight falls, all the potential energy is transferred into kinetic energy
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A 2000 kg car moves along a horizontal road at speed vo
cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
4 years ago
A 420-turn circular coil with an area of 0.0650 m2 is mounted on a rotating frame, which turns at a rate of 22.3 rad/s in the pr
Ivenika [448]

Answer:

33.48 V

Explanation:

Parameters given:

Number of turns, N = 420

Magnetic field strength, B = 0.055 T

Area, A = 0.065 m²

Angular velocity, ω = 22.3 rad/s

EMF induced in a coil is given as:

EMF = -dΦ/dt

where Φ = magnetic flux

Magnetic flux, Φ, is given as:

Φ = B * N * A * cosωt

EMF = -d( B * N * A * cosωt) / dt

EMF = B * N * A * ω * sinωt

where ωt = 90°

Therefore:

EMF = 0.055 * 420 * 0.065 * 22.3 * sin90°

EMF = 33.48 V

7 0
3 years ago
Jose is loading his luggage into his car so that he can go to visit his grandmother. He lifts his suitcase up a 10 m staircase i
maxonik [38]

Answer:

The work done on the suitcase is, W = 600 J

Explanation:

Given,

The average force exerted by Jose on his suitcase,  F = 60 N

Jose carried the suitcase to a distance, S = 10 m

The work done on the suitcase is given by the relation

                           <em>W = F x S</em>

Substituting the given values in the above equation,

                            W = 60 N x 10 m

                                 = 600 J

Hence, the work done on the suitcase is, W = 600 J

3 0
3 years ago
A 10 kg blue cart moving to the right at 25 m/s collides with a 17 kg red cart moving in the opposite direction at 16 m/s. If, a
scoray [572]

Answer:

24.8m/s

Explanation:

Given data

m1= 10kg

u1=25m/s

m2=17kg

u2=16m/s

v1=10m/s

v2=??

Applying the conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

10*25+17*16=10*10+17*v2

250+272=100+17v2

522=100+17v2

522-100=17v2

422=17v2

Divide both sides by 17

v2= 422/17

v2= 24.8 m/s

Hence the velocity of the red cart is 24.8m/s in the opposite direction of the blue cart

3 0
3 years ago
two forces 3N and 4N act on a body in a direction due north and due East respectively calculate their equivalent​
Lostsunrise [7]

Two forces 3N and 4N act on a body in a direction due north From East, the equilibrant's angle is given by \theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}.

<h3>What are equilibrium and resultant force?</h3>

The equilibrium force is the balanced force when the net force acting is zero and is the exact opposite of the consequent force. The resultant force is one single force replaced by numerous forces.

<h3>Briefing:</h3>

3N and 4N are the two forces pulling on a body.

The forces work along the North and the East, which are perpendicular to one another.

The resultant of the forces, which is provided by the equilibrant force,

R  = √(3)²+(4)²

R = 5N

From East, the equilibrant's angle is given by

\theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}

To know more about equilibrium force visit:

brainly.com/question/12582625

#SPJ9

6 0
1 year ago
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