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son4ous [18]
3 years ago
9

ANSWER ASAP

Physics
2 answers:
lidiya [134]3 years ago
5 0

Answer: The answer is B.

<em><u>Please mark me brainiest</u></em>

vodka [1.7K]3 years ago
3 0

B the desert was once covered in water

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The SI units for momentum are: Question 3 options: mph Newtons kg*m/s lbs*mph
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g a mass of 1.3 kg is pushed horizontally against a massless spring with a spring constant of 58 n/m until the spring compresses
ExtremeBDS [4]

Answer: 1.102\ J

Explanation:

Given

Mass m=1.3\ kg

Spring constant k=58\ N/m

Compression in the spring x=19.5\ cm\ or\ 0.195\ m

When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.

\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J

The kinetic energy of the mass is 1.102 J.

5 0
3 years ago
A car accelerates uniformly from rest to 24.8 m/s in 7.88 s along a level stretch of road. Ignoring friction, determine the aver
nevsk [136]

Answer:

When he weight of the car is 8.55 x 10^{3} N then power = 314.012 KW

When he weight of the car is 1.10 x 10^{4}  N then power =  43.76 KW

Explanation:

Given that

Initial velocity V_{1} = 0

Final velocity V_{2} = 24.8 \frac{m}{s}

Time = 7.88 sec

We know that power required to accelerate the car is given by

P = \frac{change \ in \ kinetic \ energy}{time}

Change in kinetic energy Δ K.E = \frac{1}{2} m (V_{2}^{2} - V_{1}^{2}   )

Since Initial velocity V_{1} = 0

⇒ Δ K.E = \frac{1}{2} m V_{2}  ^{2}

⇒ Power P = \frac{1}{2} \frac{m}{t}  V_{2} ^{2}

⇒ Power P = \frac{1}{2} \frac{W}{g\ t}  V_{2} ^{2}   -------- (1)

(a). The weight of the car is 8.55 x 10^{3} N = 8550 N

Put all the values in above formula

So power P = \frac{1}{2} \frac{8550}{(9.81)\ (7.88)} (24.8) ^{2}

P = 314.012 KW

(b). The weight of the car is 1.10 x 10^{4} N = 11000 N

Put all the values in equation (1) we get

P = \frac{1}{2} \frac{11000}{(9.81)\ (7.88)} (24.8) ^{2}

P = 43.76 KW

5 0
3 years ago
The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in 1851. Consider a l
Ket [755]

It is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Given ,

The motion of a transparent medium influences the speed of light .

The water moves with speed v in a horizontal pipe .

Assume that the light travels in the same direction as the water moves .

The speed of the light with respect to the water is c/n

Where n = 1.33 is the refractive index of water .

Let us assume ,

u' be the speed of light in water , in the frame moving with the water .

u' is related to the refractive index of water ,n as :

u'=c/n

where , c is the speed of light .

let , u be the speed of light in water in the lab frame .

Now , u and u' are related as : u = (u'+ v )/(1+ u'v/c^2)

Here v is the speed of water in the horizontal pipe .

we know the value of u' , so by substituting the value , we will get ,

u= (c/n+ v)/(1+cv/nc^2)

u= c/n(1+ nv/c)/(1+v/nc)

(b) We have , v<<c

v/c<<1 .

so , (1+v/nc )^-1 = (1-v/nc)

Now substituting this , we will get ,

u = c/n(1+nv/c) (1-v/nc)

u≈c/n(1+ nv/c-v/cn)

u≈c/n + v - v/n^2

Hence , it is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Learn more about speed here :

brainly.com/question/13943409

#SPJ4

Disclaimer : incomplete question , here is the complete question .

Question: The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc) . (b) show that for v<<c , the expression from part (a) becomes , to a good approximation , u ≈ c/n + v - v/n^2 .

8 0
1 year ago
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