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2 years ago

Why does a box on the seat of a car slide around on the c one the car speeds up slows down or turns a corner?

1 answer:

8 0

Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.

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Branches of sicence

Sonja [21]

Answer:

physical science

earth science and life science

5 0

2 years ago

Read 2 more answers

How is the rotation of the sun, different from the rotation of the earth

vesna_86 [32]

Answer:

The Sun has a north and south pole, just as the Earth does, and rotates on its axis. However, unlike Earth which rotates at all latitudes every 24 hours, the Sun rotates every 25 days at the equator and takes progressively longer to rotate at higher latitudes, up to 35 days at the poles. This is known as differential rotation.

Explanation:

6 0

2 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

2 years ago

An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is

Lemur [1.5K]

Answer:

R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

R = v₀² sin 2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

R = 9.14² sin (2 35) / 9.8

R = 8.01 m

this is the jump length

5 0

2 years ago

A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object

Tamiku [17]

Answer:

So the mass of the second object M will be 1.951 kg

Explanation:

We have given mass of the first object $m_1=3.5kg$ and its velocity $v_1=8.1m/sec$

Mass of the second object $m_2=M$ it is at rest so its velocity $v_2=0$

From conservation of momentum we know that

Initial momentum = final momentum

So $m_1v_1+m_2v_2=(m_1+m_2)v$

$3.5\times 8.1+M\times 0=(3.5+M)5.2$

$28.35=18.2+5.2M$

M = 1.951 kg

8 0

2 years ago