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3 years ago

If a man weight 155 lb on earth, specify

1 answer:

3 0

For the answer to the question above, on Earth, a one-pound object has a mass of about 0.453592 kilograms.

Therefore the man's mass is 155 * 0.453592 = 70.30676 kilograms. 

The part about the Moon's gravity is irrelevant. While the weight of a person or object would be different on the Moon, the mass would be the same.

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a train car of mass 444 kg moving at 5 m/s bounces into another car on the same tracks of mass 344 king. of the second car was m

Kamila [148]

Answer:

3.7 m/s

Explanation:

M = 444 kg

U = 5 m/s

m = 344 kg

u = - 5 m/s

Let the velocity of train is V and the car s v after the collision.

As the collision is elastic

By use of conservation of momentum

MU + mu = MV + mv

444 x 5 - 344 x 5 = 444 V + 344 v

500 = 444 V + 344 v

125 = 111 V + 86 v .... (1)

By using the formula of coefficient of restitution ( e = 1 for elastic collision)

$e = \frac{V-v}{u-U}$

-5 - 5 = V - v

V - v = - 10

v = V + 10

Substitute the value of v in equation (1)

125 = 111 V + 86 (V + 10)

125 = 197 V + 860

197 V = - 735

V = - 3.7 m/s

Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.

4 0

3 years ago

3) Ęplain why muddy water is Heterogeneous mixture

nataly862011 [7]

Answer:

muddy water is a heterogeneous mixture, which is Suspension.

7 0

3 years ago

If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of

Yuliya22 [10]

Answer:

The SI units of the “A” is m (meters)

The SI units of the “B” is m/s^2

Explanation:

Given the distance = d meters.

Time taken to travel = t (seconds)

Function of the distance, d = A + Bt^2

Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.

Thus, the SI units of the “A” is = m (meters)

The SI units of the “B” is = m/s^2

6 0

3 years ago

A car starts from rest. 900 s later it is traveling at a velocity of 30 m/s north. Which of the following equations should you u

gogolik [260]

(vx)f=(vx)i + a(t)
since it starts from rest the initial velocity is zero so you can do some algebra and get your (a).

4 0

2 years ago

A wildebeest calf is cruising at its top speed of v= 10 m/s when it passes over a sleeping cheetah. By the time the cheetah stan

Gre4nikov [31]

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

The given parameters;

speed of the wildebeest calf, Vw = 10 m/s

distance traveled by the calf before the cheetah stands up = 7 m

constant acceleration of the cheetah, a = 9.5 m/s²

Let the speed of the cheetah = Vc

let the time the cheetah catches up with the wildebeest = t

$a = \frac{V_c}{t}$

$V_c = at$

Apply relative velocity formula to determine the time when the cheetah catches up with the widebeest;

Assuming the wildebeest and the cheetah are running in the same direction;

$(V_c - V_w) t = 7 \\\\(at - V_w) t = 7\\\\(9.5t - 10)t = 7\\\\9.5t^2 - 10t = 7\\\\9.5t^2 -10t-7 = 0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 9.5 \ b= -10 \ , c =-7\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2-4ac} }{2a} \\\\t = \frac{-(-10)\ \ +/- \ \ \sqrt{(-10)^2-4(9.5\times -7)} }{2(9.5)}\\\\t = \frac{10 \ \ +/- \ \ 19.13}{19} \\\\t = 1.53 \ s$

The time when the cheetah catches up with the widebeest is 1.53 s

If the initial separation distance approaches zero;

$(V_c - V_w) t = 0\\\\(at - V_w) t = 0\\\\(9.5t - 10)t = 0\\\\9.5t^2 - 10t = 0\\\\9.5t^2 = 10t\\\\9.5t = 10\\\\t = \frac{10}{9.5} = 1.05 \ s$

Thus, if the initial separation distance approaches zero, it will take the cheetah 1.05 s to catch the widebeest.

Lear more here: brainly.com/question/24430414

6 0

3 years ago