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3 years ago

A polar bear runs at a speed of 8.3 m/s, how far will it travel in 60 seconds?

Physics

2 answers:

Kay [80]3 years ago

6 0

498 mps since you just times 8.3 by 60

Elanso [62]3 years ago

3 0

Answer:

it would have run 498 miles

Explanation:

if the polar bear runs 8.3 every second then multiply 8.3 times sixty to get 498

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The displacement in simple harmonic motion is a maximum when the 1. velocity is a maximum. 2. kinetic energy is a maximum. 3. ve

dlinn [17]

Answer:

3. velocity is zero.

Explanation:

The velocity of a simple harmonic motion is given by

$v = \omega\sqrt{A^2-x^2}$

Here, ω is the angular velocity, A is the amplitude (or maximum displacement from the equilibrium point) and x is the displacement at any time.

At maximum displacement, x = A. Then

$v = \omega\sqrt{A^2-A^2} = 0$

Therefore, at maximum displacement, velocity is 0.

Practically, this can be observed in a simple pendulum. As it approaches the maximum displacement, its velocity reduces. It becomes zero at this point and then reverses as the pendulum changes course. Then the velocity begins to increase. It becomes maximum at the equilibrium point but once past that, the velocity begins to reduce as it approaches the other amplitude.

For acceleration,

$a = -\omega^2x$

It follows that at maximum displacement, the acceleration is a maximum. The negative sign indicates that it is in an opposite direction to the displacement. Both kinetic energy ( $\frac{1}{2}mv^2$ ) and linear momentum ( $mv$ ) are proportional to velocity; they are therefore both zero at the maximum displacement.

5 0

3 years ago

A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)

from (1), (2) and (3)

$a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g$

$a = \frac{4.5-3}{3+4.5+0.75}\times 9.8$

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

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3 years ago

What kind of surface is suitable for infared radiation

nlexa [21]

Answer:

Different surfaces

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3 years ago

A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe

Levart [38]

Answer:

Coefficient of static friction will be equal to 0.642

Explanation:

We have given acceleration $a=6.3m/sec^2$