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matrenka [14]
3 years ago
12

Energy is _______ when it's moved from one object to another.

Physics
2 answers:
zzz [600]3 years ago
3 0
Energy is (transferred) when it is moved from one obect to another.

The answer is (b)
Margarita [4]3 years ago
3 0
Energy is transferred when it's moved from one object to another.

That is because energy cannot be created nor destroyed, only conducted. <span />
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Alyssa is carrying a water balloon while running down a field at a speed of 14 m/s. She tosses the water balloon forward toward
Luda [366]
From Alyssa's point of view, the water balloon is at first at rest and then gets thrown with a velocity of 23m/s. Therefore the balloon will have a speed of 23m/s for Alyssa.

At the same time, Naya is watching, and she sees the balloon at the beginning moving at a speed of 14m/s along with Alyssa, and then pushed forward of other 23m/s. Therefore, from her point of view, the balloon will have a speed of 14+23 = 37m/s.

Hence, the correct answer is <span>D) The speed of the balloon is 23 m/s for Alyssa and 37 m/s for Naya. </span>
4 0
3 years ago
Read 2 more answers
What is the mass of a cart that accelerates at 3.0 m/s if a 0.5 N force is applied?
frozen [14]

The mass of the cart is 0.167 kg

Explanation:

We can solve the problem by using Newton's second law, which states that the net force applied to an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the net force acting on an object

m is its mass

a is its acceleration

For the cart in this problem, we have

F = 0.5 N is the net force applied

a=3.0 m/s^2 is the acceleration

Substituting, we find the mass:

m=\frac{F}{a}=\frac{0.5}{3.0}=0.167 kg

Learn more about Newton's second law:

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#LearnwithBrainly

6 0
3 years ago
At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assum
Verdich [7]

Answer:

Ec = 6220.56 kcal

Explanation:

In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.

You use the following formula:

W_c=Mgh        (1)

Wc: work done by the climber

g: gravitational constant = 9.8 m/s^2

M: mass of the climber = 78.4 kg

h: height reached by the climber = 5.42km = 5420 m

You replace in the equation (1):

W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J     (2)

Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:

0.16(E_c)=4,164,294.4J

Ec: Calories

You solve for Ec and convert the result to Cal:

E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal

The amount of Calories needed by the climber was 6220.56 kcal

4 0
3 years ago
Desiree has a scientific question that she wants to investigate. She has developed a hypothesis and method for her experiment. W
avanturin [10]
The next scientific step is to collect data to test her Hypothesis. 
8 0
4 years ago
Read 2 more answers
In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten
sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

3 0
3 years ago
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