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Veronika [31]
3 years ago
7

Physics help please!

Physics
1 answer:
Oliga [24]3 years ago
6 0

Answer:

student attach a save block to a horizontal spring so that the block spring system will oscillator with the block spring system released from rest horizontal position that is not the systems equilibrium position well this question regards about the energy used the answer may be 0.73 Joel ok you just try it ok verified

Explanation:

apply applied the potential energy value mean the formula MGH write it means what mass into gravitation in to height

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Answer:

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Explanation:

4 0
3 years ago
A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope
tangare [24]

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts  in the horizontal direction can be expressed as:

F_x = F \ cos \theta

F_x = 100 \ cos (53)

F_x = 60.18 \ N

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

F_x = F \ cos \theta

F_x = 100 \ cos (180)

= 100 × -1

= -100 N

3 0
3 years ago
Sarah, who has a mass of 55 kg, is riding in a car at 20 m/s. She sees a cat crossing the street and slams on the brakes! Her se
avanturin [10]

Answer:

-2200 N

Explanation:

The change in momentum of Sarah is equal to the impulse, which is the product between the force exerted by the seatbelt on Sarah and the time during which the force is applied:

\Delta p=I\\m \Delta v = F \Delta t

where

m is the mass

\Delta v is the change in velocity

F is the average force

\Delta t is the duration of the collision

In this problem:, we have:

m = 55 kg is Sarah's mass

\Delta v = 0-20 = -20 m/s  is the change in velocity

\Delta t = 0.5 s  is the duration of the collision

Solving for F, we find the force exerted by the seatbelt on Sarah:

F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N

Where the negative sign indicates that the direction of the force is opposite to that of Sarah's initial velocity.

5 0
3 years ago
A 22.0 kg child slides down a slide that makes a 37.0° angle with the horizontal. (a) What is the magnitude of the normal force
patriot [66]

Answer:

(a) 172.185 N

(b) 53^{\circ}

Solution:

As per the question:

Mass of the child, m = 22.0 kg

Angle, \theta = 37.0^{\circ}

Now,

(a) The magnitude of the normal force exerted by the slide on the child:

F_{N} = mgcos\theta

F_{N} = 22\times 9.8cos37^{\circ} = 172.185\ N

Now,

(b) The angle from the horizontal at which the force is directed is:

90^{\circ} - 37^{\circ} = 53^{\circ}

6 0
3 years ago
A 6.00-kg box is sliding to the right across the horizontal floor of an elevator. The coefficient of kinetic friction between th
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Answer: Kinetic frictional force = 23.76N

Explanation: Please see the attachments below

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