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garik1379 [7]
3 years ago
14

A man pushes with a force of 300 N on a bookcase of mass 75 kg. The coefficient of static friction between the bookcase and the

floor is 0.9. Because of friction, the bookcase does not move. What is the force of friction on the bookcase?
A. 362 N
B. 300 N
C. 551 N
D. 662 N
Physics
2 answers:
ad-work [718]3 years ago
6 0
A! It’s A. A. It’s a. A
kondaur [170]3 years ago
4 0
362 N is the force of friction on the bookcase.
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Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?
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3 years ago
Which of the following phenomena suggest that light may be transverse wave
aniked [119]

Answer:

Polarization.

Reason being that phenomena verifies the transverse nature of light

4 0
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Astronauts on the first trip to Mars take along a pendulumthat has a period on earth of 1.50 {\rm s}. The period on Mars turns o
Fynjy0 [20]

Answer:

3.7 m/s^2

Explanation:

The period of a pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum and g is the free-fall acceleration on the planet.

In this problem, we know that the period of the pendulum on Earth is:

T_e = 1.50 s

while the period of the same pendulum on Mars is

T_m = 2.45 s

And since the length of the pendulum L does not change, we can write:

\frac{T_e}{T_m}=\frac{2\pi \sqrt{\frac{L}{g_e}}}{2\pi \sqrt{\frac{L}{g_m}}}=\sqrt{\frac{g_m}{g_e}}

where

g_e = 9.8 m/s^2 is the free-fall acceleration on Earth

g_m = ? is the free-fall acceleration on Mars

Re-arranging the equation and substituting numbers, we find:

g_m = \frac{T_e^2}{T_m^2}g_e=\frac{(1.50 s)^2}{(2.45 s)^2}(9.8 m/s^2)=3.7 m/s^2

7 0
3 years ago
The heater element of a 120 V toaster is a 5.4 m length of nichrome wire, whose diameter is 0.48 mm. The resistivity of nichrome
Ksenya-84 [330]

Answer:

The power drawn by the toaster is closest to:

(A) 370 W

Explanation:

First we calculate the resistance of the nichrome wire (R).

R=\frac{pL}{A} =\frac{pL}{\pi r^{2} } \\

Where radious (r), resistance coefficient (p), and Length (L)

r=\frac{0.48}{2} 10^{-3} m\\\\L=5.4 m\\p=1.3*10^{-3}\varOmega  \\\\\\Replacing:\\\\R=\frac{1.3(10)^{-6} *5.4}{\pi* 0.24^{2} (10)^{-6}} =38.7940

After replace the value in the ohm law power formula to obtain the power consumed:

P=\frac{V^2}{R} =\frac{120^2}{38.7940} =371.191 Watts

7 0
3 years ago
What is the total magnification for each lens setting on a microscope with 15x oculars and 4x, 10x, 45x, and 97x objectives lens
KengaRu [80]

Answer:

M = 60x

M = 150x

M = 675x

M = 1455 x

Explanation:

In a microscope the total magnification is found by multiplying the magnification of the ocular (eye piece) and objective lens.

m_e=Ocular magnification = 15x

m_o=Objective magnification = 4x

Total magnification

M=m_e\times m_o\\\Rightarrow M=15\times 4\\\Rightarrow M=60x

M = 60x

M=m_e\times m_o\\\Rightarrow M=15\times 10\\\Rightarrow M=150x

M = 150x

M=m_e\times m_o\\\Rightarrow M=15\times 45\\\Rightarrow M=675x

M = 675x

M=m_e\times m_o\\\Rightarrow M=15\times 97\\\Rightarrow M=1455x

M = 1455 x

7 0
3 years ago
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