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WINSTONCH [101]
2 years ago
12

A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in

Physics
1 answer:
KatRina [158]2 years ago
6 0

Answer:

37.5 seconds

Explanation:

The given parameters are;

The mass of the block on the spring, m = 2.50 kg

The force with which the loaded spring launches the block, F = 450 N

The change in momentum of the block, Δp = 12.0 kg·m/s

We have;

Let the force with which the block was launched = The net force, F_{NET}

By Newton's second law of motion, we have;

F = F_{NET} = Δp × Δt

Where;

Δt = The time the block is in contact with the spring

Therefore;

\Delta t = \dfrac{F_{NET}}{\Delta p}

By plugging in the values for F_{NET} and Δp, we have;

\Delta t = \dfrac{450 \ N}{12.0 \ kg \cdot m/s} = 37.5 \ s

The time duration the block is in contact with the spring, Δt = 37.5 seconds.

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3 years ago
How will creat thunderstrom​
emmasim [6.3K]

Answer:

the air has to be unstable as well as it needs to be moved upwards.

Explanation:

it needs to be moved upwards and also needs to have unstable air.

3 0
2 years ago
A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W
dalvyx [7]

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

F_g = mg

now the acceleration of ball is given as

a = \frac{F_g}{m}

a = \frac{mg}{m} = g

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

5 0
2 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
3 years ago
I have a hot air balloon that has 18.0 g helium gas inside of it. if the pressure is 2.00 atm and the temperature is 297k, what
Anastaziya [24]

M = molar mass of the helium gas = 4.0 g/mol

m = mass of the gas given = 18.0 g

n = number of moles of the gas

number of moles of the gas is given as

n = m/M

n = 18.0/4.0

n = 4.5 moles

P = pressure = 2.00 atm = 2.00 x 101325 Pa = 202650 Pa

V = Volume of balloon = ?

T = temperature = 297 K

R = universal gas constant = 8.314

Using the ideal gas equation

P V = n R T

(202650) V = (4.5) (8.314) (297)

V = 0.055 m³

4 0
2 years ago
Read 2 more answers
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