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mars1129 [50]
4 years ago
13

Compared to the earth would the sun appear to move more quickly across Mercury’s sky or more slowly? Why?

Physics
1 answer:
denis23 [38]4 years ago
7 0

Faster because of the distance mercury in to the sun

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A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is
andriy [413]

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

4 0
3 years ago
What is rotation and revolution
skad [1K]
When an object turns around an internal axis (like the Earth turns around its axis) it is called a rotation. When an object circles an external axis (like the Earth circles the sun) it is called a revolution.
5 0
3 years ago
If the road becomes wet or crowded, you should ____. slow down and increase your following distance All choices are incorrect. m
Vaselesa [24]

Answer:

The first one.

7 0
3 years ago
Which statements accurately describe Dmitri Mendeleev’s contributions to the development of the periodic table? Check all that a
Levart [38]
He produced the first orderly arrangement of known elements, he used patterns to predict undiscovered elements
6 0
3 years ago
Read 2 more answers
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
3 years ago
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