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Monica [59]
3 years ago
15

The diagram shows the apparent motion of the Sun across the sky. EAST WEST Sunrise Which action causes the Sun to appear to move

in this way?
A. Earth rotates from south to north.

B. Earth rotates from north to south.

C. Earth rotates from west to east.

D. Earth rotates from east to west.​
Physics
2 answers:
Andreas93 [3]3 years ago
8 0
C. West to east is the correct answer
ra1l [238]3 years ago
8 0
C is the right answer
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A student pushes a 50-N box across the floor a distance of 15 m. How much work was done to move the box?
irinina [24]

Answer:750

Explanation:

50 times 15

4 0
2 years ago
Read 2 more answers
An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos
erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v}    \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm

3 0
3 years ago
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
Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0
lora16 [44]

Answer:

<h2>154.73N</h2>

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

Ty = 7.73N

W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

3 0
3 years ago
A gas is compressed in a cylinder from a volume of 20.0 l to 2.0 l by a constant pressure of 10.0 atm. calculate the amount of w
jok3333 [9.3K]
First, we need to convert the pressure in SI units. Keeping in mind that 1 atm = 1.01 \cdot 10^5 Pa:
p=10.0 atm =1.01 \cdot 10^6 Pa

The initial and final volume of the gas are (keeping in mind that 1.0 L = 0.001 m^3):
V_i = 20.0 L=0.020 m^3
V_f = 2.0 L=0.002 m^3

So, the work done on the gas by the surrounding is
W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J
And the final positive sign means that this work corresponds to an increase in internal energy of the gas.
8 0
3 years ago
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