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GarryVolchara [31]

3 years ago

8

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s

starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared.

1 answer:

Doss [256]3 years ago

5 0

<h2>Acceleration of cheetah is 10 m/s²</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 50 miles/hr = 80 km/hr = 22.22 m/s

Time, t = 2.22 s

Substituting

v = u + at

22.22 = 0 + a x 2.22

a = 10 m/s²

Acceleration is 10 m/s²

Acceleration of cheetah is 10 m/s²

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OlgaM077 [116]

Answer:

The speed with which the man flies forward is 5.5 m/s

Explanation:

The mass of the man = 100 kg

The mass of the scooter = 10 kg

The speed with which the man was traveling on the scooter = 5 m/s

The speed of the scooter after it hits the rock = 0 m/s

Let v represent the speed with which the man flies forward

The formula for momentum, P, is P = Mass × Velocity

The conservation of linear momentum principle is, the total initial momentum = The total final momentum, therefore, we have;

The total initial momentum = (100 kg + 10 kg) × 5 m/s = 550 kg·m/s

The total final momentum = 100 kg × v + 10 kg × 0 m/s = 100 kg × v

When the momentum is conserved, we have;

550 kg·m/s = 100 kg × v

∴ v = 550 kg·m/s/(100 kg) = 5.5 m/s.

The speed with which the man flies forward = v = 5.5 m/s

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2 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

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The correct one is A. Technician Only.
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How many stars are on the us flag?

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A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

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3 years ago

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