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2 years ago

Acceleration Practice

Physics

2 answers:

Trava [24]2 years ago

8 0

Answer:

0.85 m/s²

Explanation:

Acceleration is change in velocity over change in time. In mathematically, it can be expressed as:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec v_2 - \vec v_1}{t_2-t_1}}$

Our final velocity is given to be 17 m/s in 20 seconds. Initial velocity is at starting point which is 0 m/s in 0 second. Therefore:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17-0}{20-0}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17}{20}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = 0.85 \ \, \sf{m/s^2}}$

Therefore, the acceleration of a horse from starting point to 17 m/s in 20 seconds is 0.85 m/s²

just olya [345]2 years ago

4 0

Answer:

0.85 m/s^2

Explanation:

Acceleration = (final velocity - starting velocity) / time

Final velocity = 17

Starting velocity = 0

Time = 20

1. Substitute:

Acceleration = (starting velocity - final velocity) / time --> Acceleration = ( 17 - 0) / 20

2. Solve:

Acceleration = ( 17 - 0) / 20 -->

Acceleration = 0.85

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A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

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Answer:

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Explanation:

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A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

Softa [21]

Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

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