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bekas [8.4K]
3 years ago
10

At a college homecoming 20 students lifted a sports car. Each student exerted 450N of upward force on the car. a) What is the ma

ss of the car in kg? b) What is its weight in lbs?
Physics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

(a) 918.37 kg

(b) 2020.41 lbs

Explanation:

number of students = 20

Force of each student = 450 N

Total upward force = 450 x 20 = 9000 N

(a)

Let m be the mass of car

F = m x g

9000 = m x 9.8

m = 918.37 kg

(b)

As we know that, 1 kg = 2.20 lbs

So, 918.37 kg = 2.20 x 918.37 = 2020.41 lbs

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A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista
Nastasia [14]

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

8 0
3 years ago
In a real world trampoline, how do gymnasts keep a constant bounce height
kkurt [141]
By using common factors of physics: weight, gravity, and stability. 
Weight would keep them at a constant height.
Gravity helps the weight and how much force it propels the person, or objects, into the air.
Stability helps adjust how much distance the person, or object, needs to be.
7 0
3 years ago
A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i
sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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