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dybincka [34]
3 years ago
14

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

m high. How far from the base of the cliff does it land?
Physics
1 answer:
Vikki [24]3 years ago
3 0

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


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The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a
Setler [38]

Answer:

 x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

         x_{cm} = 1 / M_{total}  ∑ x_{i}  m_{i}

Where M_{total} is the total masses of the system, x_{i} is the distance between the particles and m_{i} is the masses of each body

Let's apply this equation to our problem

        M = Me + m

        M = 5.98 10²⁴ + 7.36 10²²

        M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

       x_{cm} = 1 / 605.36 10²²   [Me 0 + 7.36 10²² 3.82 10⁸]

       x_{cm} = 4.644 10⁶ m

4 0
3 years ago
I need help answering 30A and B. Help a guy out?
dmitriy555 [2]

Uh so I'm no master at this subject, but all stuffs accelerate at 9.8 m/s squared. So you multiply the 9.8 and the 0.20 it's given for reasons unknown other than that's what I see in my notes... and that gives you 1.96 m/s squared.

As for B, I have no idea. I think you may multiply the 1.96 by 4. Tell me your thoughts and maybe we can work it out together

5 0
3 years ago
Question 2 (10 points)
kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

F=ma

where

m = 1390 kg is the mass of the truck

a=-3.6 m/s^2 is the acceleration

And substituting,

F=(1390)(-3.6)=-5004 N

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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