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kramer
3 years ago
10

A car of mass 998 kilograms moving in the positive y–axis at a speed of 20 meters/second collides on ice with another car of mas

s 1,200 kilograms moving in the negative x direction with a speed of 17 meters/second. After the collision, the cars stick and move together. What is the total momentum in the x direction before the collision?
Physics
1 answer:
goldfiish [28.3K]3 years ago
3 0
    <span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400

This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.

Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.

Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600

This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.

Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)

The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.

x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
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You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100
Vikki [24]

m = mass of the cart = 85 kg

F = net force on the cart = 250 N

a = acceleration of the cart

acceleration of the cart is given as

a = F/m

a = 250/85

a = 2.94 m/s²

t = time for which the force is applied = 5 sec

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart just before  it flies off the cliff = ?

using the equation

v = v₀ + a t

inserting the values

v = 0 + (2.94) (5)

v = 14.7 m/s

consider the motion of cart after it flies off the cliff in vertical direction :

v' = initial velocity in vertical direction = 0 m/s

a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²

t' = time taken for the cart to land = ?

Y' = vertical displacement of the cart = height of cliff = 100 m

using the kinematics equation

Y' = v' t' + (0.5) a' t'²

100 = (0) t' + (0.5) (9.8) t'²

t' = 4.52 sec


consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

t' = time of travel = 4.52 sec

v = velocity along the horizontal direction = 14.7 m/s

distance traveled from the base of cliff is given as

X = v t'

X = 14.7 x 4.52

X  = 66.4 m


3 0
3 years ago
a man of mass 50 kg climbs up stairs each of height 0.2 m in 20 seconds .calculate the power of the man​
Delicious77 [7]
How many stairs?
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Gala2k [10]

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2 years ago
A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v
hram777 [196]

Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

v is the final velocity = 30m/s

u is the initial velocity = 20m/s

t is the time = 5secs

m = 300kg

Get the net force:

Recall that: F = m(v-u)/t

F  = 300(30-20)/5

F = 60(30-20)

F = 60(10)

<em>F = 600N</em>

<em>Hence the net force acting on the car is 600N.</em>

<em></em>

<em></em>

3 0
3 years ago
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