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Delvig [45]
3 years ago
11

A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds

of its travel. What is its average speed during the time from t = 5 seconds to t = 10 seconds?Select one of the options below as your answer:. A. 2 meters/second. B. 3 meters/second. C. 4 meters/second. D. 5 meters/second. E. 6 meters/second . will hand out medals
Physics
2 answers:
igomit [66]3 years ago
8 0
The total distance traveled by the body is the sum of the distance it traveled for the first five seconds which is 10 m and that of the next five seconds which is 30 meters. Thus, the total distance traveled is 40 m. Dividing this by the time, will give the average speed. The average speed is therefore, 4 m/s. The answer is letter C. 
natita [175]3 years ago
7 0

Answer:

c

Explanation:

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The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t
Vika [28.1K]

Answer:

The hydro static force on the back of the dam is 1.96\times10^{11}\ N

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

P_{avg}=\rho\times g\times d_{avg}

Put the value into the formula

P_{avg}=1000\times9.8\times100

P_{avg}=980000\ Pa

We need to calculate  the hydro static force on the back of the dam

Using formula of force

F = P_{avg}\times A

Put the value into the formula

F = 980000\times1000\times200

F=1.96\times10^{11}\ N

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7 0
3 years ago
Which title best reflects the main idea of the passage? The Role of Convection in the Distribution of Earth's Energy The Role of
Leto [7]

Answer:

The Role of Heat Transfer Methods in the Distribution of Earth's Energy

Explanation:

8 0
3 years ago
Read 2 more answers
Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface
mina [271]
We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = \frac{1}{2} m*v^2
GPE (Gravitational Potential Energy) = m*g*h

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
TE =  \frac{1}{2} m*v^2 + m*g*h

Solving:
TE = \frac{1}{2} m*v^2 + m*g*h
TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50
TE =  \frac{2.0*100}{2} + 1000
TE =  \frac{200}{2} + 1000
TE = 100 + 1000
\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark


5 0
2 years ago
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A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m
Nina [5.8K]

Answer: 0.05\ mm

Explanation:

Given

Cross-sectional area of wire A_1=4\ mm^2

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Extension in a wire is given by

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\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

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What happens with alpha, beta, and gamma radiation?
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I'm sorry but I don't really understand the question. What is the quest actually asking???
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