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3 years ago

A girl lifts a 160-N load to a height of 1 min 0.5 s. How much power is used to lift the load?

Physics

1 answer:

Aleonysh [2.5K]3 years ago

8 0

Answer: 320 W

Explanation:

P = W/t

W = F * d

F = 160 N

d = 1 m

t = 0.5 s

W = 160 * 1

W = 160 J

P = 160 J / 0.5 s

P = 320 W

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A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

GrogVix [38]

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction ( $\mu$ )=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

$v^2-u^2=2as$

$acceleration=\mu g=0.8\times 9.8=7.84 m/s^2$

s=2.4 m

$0-(u)^2=2\times (-7.84)\times 2.4$

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

$920\times v=(920+2300)\times 6.13$ (conserving momentum)

v=21.45 m/s

7 0

3 years ago

A ray of yellow light ( f8= 5.09 × 1014 Hz) travels at a speed of 2.04×10 meters per second in

Gala2k [10]

The complete question in the attached figure
Let
c ------------------- > is the speed of light
v ------------------- > is the speed in medium
n ------------------- > is the refractive index of medium
we know that
c/v = n
n = (3 x 10^8)/(2.04 x 10^8) = 1.47

1.47 is the refractive index of glycerol.
therefore

the answer is (4) glycerol

5 0

3 years ago

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An object is thrown off a cliff with a horizontal speed of 10 m/sec. After 3 seconds the object hits the ground. Find the height

Degger [83]

Answer:

223

Explanation:

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3 years ago

The compressed-air tank ab has a 250-mm outside diameter and an 8-mm wall thickness. it is fitted with a collar by which a 40-kn

valentinak56 [21]

<span>Assume: neglect of the collar dimensions. Ď_h=(P*r)/t=(5*125)/8=78.125 MPa ,Ď_a=Ď_h/2=39 MPa Ď„=(S*Q)/(I*b)=(40*ă€–10ă€—^3*Ď€(ă€–0.125ă€—^2-ă€–0.117ă€—^2 )*121*ă€–10ă€—^(-3))/(Ď€/2 (ă€–0.125ă€—^4-ă€–0.117ă€—^4 )*8*ă€–10ă€—^(-3) )=41.277 MPa @ Point K: Ď_z=(+M*c)/I=(40*0.6*121*ă€–10ă€—^(-3))/(8.914*ă€–10ă€—^(-5) )=32.6 MPa Using Mohr Circle: Ď_max=(Ď_h+Ď_a)/2+âš(Ď„^2+((Ď_h-Ď_a)/2)^2 ) Ď_max=104.2 MPa, Ď„_max=45.62 MPa</span>

3 0

3 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago