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Sati [7]
3 years ago
6

11) A sled is initially given a shove up a frictionless 35º incline. It reaches a maximum height of 2.5

Physics
1 answer:
Andrej [43]3 years ago
5 0

Answer:

7 m/s

Explanation:

To solve this problem you must use the conservation of energy.

K1 +U1=K2+U2

That math speak for, initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.

The initial PE (potential energy) is 0 because it hasn't been raised in the air yet. The final KE (kinetic energy) is 0 because it isn't moving. This gives the following:

KE1= \frac{1}{2}mv^{2}}

PE2=mgh

K1=U2

\frac{1}{2} mv^{2} =mgh

Solve for v

v=\sqrt{2gh}

Input known values and you get 7 m/s.

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A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
10.
nlexa [21]

Answer:

Cell Membrane

Explanation:

The cell membrane controls what goes in and out of a cell, and keeps it shape, much like a city limit.

3 0
3 years ago
Charina says that when waves interact with an object, they will interfere with the object, and when waves interact with other wa
Fofino [41]
I would not agree with her since reflection and refraction happens only when waves hit an object. When, waves meet it is either it experiences constructive or destructive interference. Hope this answers the question. Have a nice day.
3 0
3 years ago
Read 2 more answers
calculate how much charge passes through the LED in 1 hour if the current flowing in the circuit below is 350mA
Morgarella [4.7K]

Given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

<h3>What is Current?</h3>

Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.

If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;

I = Q/t

Where Q is the charge and t is time elapsed.

Given the data in the question;

  • Time elapsed t = 1hr = 3600s
  • Current I = 350mA = 0.35A
  • Charge Q = ?

We substitute our given values into the expression above to determine the charge.

I = Q/t

Q = I × t

Q = 0.35A × 3600s

Q = 1260C

Therefore,  given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

Learn more about current here: brainly.com/question/3192435

#SPJ1

3 0
2 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit
Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7\times 10^{10} m

v_i=9.5\times 10^4 m/s

d_2=6\times 10^{12} m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}

\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f

v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})

v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}

Using the formula

v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}

Where mass of sun=M=1.98\times 10^{30} kg

G=6.7\times 10^{-11}

v_f=58515.9 m/s

4 0
3 years ago
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