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mihalych1998 [28]
3 years ago
13

Will water pressure be greater at the bottom of a container that you put a hole in or at the top of a container?

Physics
1 answer:
ddd [48]3 years ago
4 0
Pressure=hrg
pressure is great at the bottom because the weight lf water exerts pressure below the container due to the gravitational pull of the earth.
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Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr
Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0
3 years ago
Read 2 more answers
A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a
klemol [59]

Answer:

15.4 kg.

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

Let: u = x m/s, and V = 1/3x m/s

Substitute into equation 1

7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

7 0
3 years ago
Read 2 more answers
Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}

T_2=266.12K

Therefore, the final temperature of gas is 266.12 K

5 0
3 years ago
In which material does the light wave have the larger wavelength?
pochemuha

Answer: in the smaller/ thinner pieces.

8 0
2 years ago
A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes
Lerok [7]

Answer:

R = 715.4 N

L =  166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L =  166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

                                             0 = 0  checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

5 0
3 years ago
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