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SVETLANKA909090 [29]
4 years ago
10

How much power is needed to push a 95 kg chest at 0.67 m/s along a horizontal floor where the coefficient of friction is 0.77?

Physics
1 answer:
Savatey [412]4 years ago
8 0
F = mg\mu \\ \\  P =  \frac{Fdx}{dt} = Fv  \\ \\  P =  mg\mu\ v
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Answer:

The answer is 1.0 N

Explanation:

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta  is parallel to the tray =1.039 N

5\cos \theta  is perpendicular to the tray =4.890 N

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Answer:

B

Explanation:

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7 0
3 years ago
The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal
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Α=2.4 \frac{m}{s}

β=1.2 \frac{m}{s^2}

x(t)=at

y(t)=3-βt^2

Vx(t)=α

Vy(t)=-2βt

vectorV=[α;-2βt]

ax(t)=0

ay(t)=-2βt

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3 0
4 years ago
Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
Nezavi [6.7K]

Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0  x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12    (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))  

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR)  --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides  

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x 10^{-9  

t = 23.9ns

Thus, it took 23.9ns  for the potential difference between the deflection plates to reach 75 volts

6 0
3 years ago
Read 2 more answers
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