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2 years ago

Acceleration is a change in speed or direction over time. In what two ways does the sled accelerate as it descends?

Physics

1 answer:

Anna007 [38]2 years ago

4 0

Answer:

By force and sloppy surface.

Explanation:

By applying force on the sled and slope of the path are the two ways the sled accelerate as it descends. If there is more friction between sled and the ground then force is required to push the sled to move downward while on the other hand, if the path on which sled moves is sloppy then it will move automatically without the use of force, so these two ways can accelerate the sled.

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How do you think the switch controls the flow of current to the light?

nadezda [96]

Answer:

A switch interrupts the flow of current in a circuit.

Explanation:

3 0

2 years ago

A roller coaster is moving at 25m/s at the bottom of a hill. Three seconds later it reaches the top of the hill moving at 10m/s.

marishachu [46]

Answer:

$5 m/s^{2}$

Explanation:

acceleration is defined as the rate of change of velocity per unit time. Therefore, acceleration, $a=\frac {\triangle v}{\triangle t}$

where $\triangle v$ is change in velocity and t is time

Substituting initial and final velocities with 25 m/s and 10 m/s then using time as given in the question of 3 s then

$a=\frac {25-10}{3}=\frac {15}{3}=5 m/s^{2}$

Therefore, the acceleration is $5 m/s^{2}$

8 0

3 years ago

A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

$v=3.49 m/s$

7 0

3 years ago

a stone is dropped from rest at an initial height h above the surface of the earth. Show that the speed with which it strikes th

Ksju [112]

Answer:

$v=\sqrt{2gh}$

Explanation:

We could use conversation of energy. Total distance the stone will cover will be

$h=\frac{1}{2} g t^{2}$

the Final velocity will be

$v=gt\\v=g\sqrt{\frac{2h}{g} }\\ v=\sqrt{2gh}$

5 0

3 years ago

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, $P_{gauge \ pressure}$ = 2.95 atm

Atmospheric pressure, $P_{atm}$ = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2$

where;

P₁ = $P_{gauge \ pressure} + P_{atm \ pressure}$

P₂ = $P_{atm}$

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }$

where;

$\rho$ is the density of seawater = 1030 kg/m³

$v_2 = \sqrt{ \frac{2(2.95*101325 \ + \ 1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s$

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0

3 years ago