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Bumek [7]
3 years ago
14

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

the cart and the velocity of the cart before it hits the wall. What other measurement is necessary to calculate the impulse delivered to the lab cart?
the velocity of the cart after it hits the wall
the force exerted on the cart by the wall
the time the cart is in contact with the wall
the acceleration of the cart before it hits the wall
Physics
2 answers:
Rzqust [24]3 years ago
5 0

Answer: the velocity of the cart after it hits the wall

Explanation:

Impulse is the force exerted in a small interval of time. It can also be defined as the change in momentum.

I = F Δt = m v - m u = m(v-u)

where, m is the mass of the object, v is the final velocity and u is the initial velocity.

It is measured in Newton-seconds or kilogram-meter per second.

When the mass of the cart and initial velocity is given, the final velocity is required after it hits the wall in order to calculate the impulse delivered to the lab cart.

WINSTONCH [101]3 years ago
4 0

The correct answer to the question is : A) The velocity of the cart after it hits the wall.

EXPLANATION:

Before answering this question, first we have to understand impulse.

Impulse of a body is defined as the change in momentum or the product of force with time.

Mathematically impulse = m ( v- u ).

Here, v is the final momentum and u is the initial momentum.

Hence, we need the velocity of the cart after it hits the wall in order to calculate the impulse of the lab cart.

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tatyana61 [14]

Answer:

D More: Less

Explanation:

5 0
2 years ago
U. A hockey player takes a slap shot at a puck at rest on
faltersainse [42]

Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s².  Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

(c) The time the puck takes to travel 10 ft without friction is:

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s

8 0
2 years ago
A science class puts a balloon containing 1.25 l of air at 101 kpa into a bell jar. using an air pump, the class removes some of
storchak [24]
The first thing you should know in this case is the following definition:
 PV = nRT
 Then, as the temperature is constant, then:
 PV = k
 Then, we have two states:
 P1V1 = k
 P2V2 = k
 We can then equalize both equations:
 P1V1 = P2V2
 Substituting the values:
 (1.25) * (101) = (2.25) * (P2)
 Clearing P2:
 P2 = ((1.25) * (101)) /(2.25)=56.11Kpa
 answer:
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8 0
3 years ago
When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d
Alchen [17]

Answer:

g = 0.85 ms^{-2}

Explanation:

g = \frac{GM}{h^{2} }

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 10^{-11} Nm^{2}kg^{-2}), M is the mass of the earth (5.972 x 10^{24} kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \frac{6.674*10^{-11}*5.972*10^{24}  }{(21661400)^{2} }

  = \frac{3.9857 *10^{14} }{4.6922*10^{14} }

  = 0.84944

g = 0.85 ms^{-2}

The acceleration due to the Earth's gravitation is 0.85 ms^{-2}.

6 0
2 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
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