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s2008m [1.1K]
3 years ago
13

Edwin Hubble is most famous for?

Physics
1 answer:
Aleks04 [339]3 years ago
4 0
I believe it is the answer is A
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Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
The diagram shows a person using a piece of gym equipment to lift weights.
ki77a [65]

Answer:

C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

Explanation:

3 0
3 years ago
Read 2 more answers
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

PV=nRT

For a gas

P_{1}V_{1}=nRT_{1}

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

10\times V=nR\times286....(I)

When the temperature of the gas is increased

Then,

P_{2}V_{2}=\dfrac{n}{2}RT_{2}....(II)

Divided equation (I) by equation (II)

\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}

\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}

P_{2}=\dfrac{10\times368}{2\times286}

P_{2}= 6.433\ atm

P_{2}=6.4\ atm

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0
3 years ago
Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up
Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

v = \sqrt{2as}

v = \sqrt{2*18.9*391}

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

V=  \sqrt{v^2 + 2aS'}

V = \sqrt{121.57^2+ 2 (-9.92)(332)}

V = 90.51 m/s

6 0
3 years ago
A mover pushes a 46.0kg crate 10.3m across a rough floor without acceleration. How much work did the mover do (horizontally) pus
Alchen [17]

Answer:

<h3>2,321.62Joules</h3>

Explanation:

The formula for calculating workdone is expressed as;

Workdone = Force * Distance

Get the force

F = nR

n is the coefficient of friction = 0.5

R is the reaction = mg

R = 46 ( 9.8)

R = 450.8N

F = 0.5 * 450.8

F = 225.4N

Distance = 10.3m

Get the workdone

Workdone = 225.4 * 10.3

Workdone  = 2,321.62Joules

<em>Hence the amount of work done is 2,321.62Joules</em>

3 0
3 years ago
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