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1 year ago

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A 8.0 n force acts on a 0.70-kg object for 0.50 seconds. by how much does the object's momentum change (in kg-m/s)? (never inclu

de units in the answer to a numerical question.)

1 answer:

inysia [295]1 year ago

4 0

For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
Rewriting the acceleration as the increment of velocity $\Delta v$ in a time $\Delta t$ : $a= \frac{\Delta v}{\Delta t}$ , F becomes
$F=m \frac{\Delta v}{\Delta t}$
But given the definition of momentum: $p=mv$ , then $m \Delta v$ represents the momentum change. So we can rewrite F as
$F= \frac{\Delta p}{\Delta t}$
And re-arranging the formula we can calculate the value of the change in momentum:
$\Delta p = F \Delta t=(8.0 N)(0.50 s)=4 kg m/s$

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kozerog [31]

Work = Force x Distance
47.2J = 23.3N x d
d = 47.2/23.3
d = 2.0258 m

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1 year ago

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A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

svp [43]

Answer:

The kinetic energy of the clam at a height of 5.0 m is 5.19 J and the speed of the clam at that height is 9.71 m/s.

<u>Explanation: </u>

<em>Mechanical energy is constant throughout the travel</em>, we know that <em>mechanical energy is calculated by adding potential energy and kinetic energy</em>. Potential energy = $m \times g \times h$ , Kinetic energy = $\frac{1}{2} \times m \times v^{2}$ and Mechanical energy = $m \times g \times h+\frac{1}{2} \times m \times v^{2}$ Kinetic energy is zero at initial point. Now mechanical energy of clam with m=0.11kg,g=9.81 $\frac{m}{s^{2}}$ ,h=9.8 m is = 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of clam at a height of 5.0 m = $0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2}$ = $5.39+\frac{1}{2} \times m \times v^{2}$ . We know that mechanical energy is constant hence, <em>mechanical energy of clam at height 9.8 m is equal to mechanical energy at height 5.0 m</em>. This is represented as following

10.58 = $5.39+\frac{1}{2} \times m \times v^{2}$ 10.58 – 5.39 = $\frac{1}{2} \times m \times v^{2}$ 5.19 = $\frac{1}{2} \times m \times v^{2}$ kinetic energy of the clam is 5.19 J.

Now speed of the clam at height 5.0 m is 5.19 = $\frac{1}{2} \times 0.11 \times v^{2} \frac{5.19 \times 2}{0.11}=v^{2}$ 94.36 = $v^{2} \sqrt{94.36}=v \quad v$ = 9.71 m/s. The speed of the clam is 9.71 m/s.

6 0

1 year ago

A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

#learnwithBrainly

8 0

1 year ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

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2 years ago