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Cerrena [4.2K]
2 years ago
6

A constant braking force of 10 newtons applied for 5 seconds is used to stop a 2.5-kilogram cart traveling at 20 meters per seco

nd. the magnitude of the impulse applied to stop the cart is
Physics
2 answers:
ss7ja [257]2 years ago
6 0
Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
 The magnitude of the impulse applied to stop the cart is
 I = FpΔt = (10N) * (5s) = 50 N.s
inn [45]2 years ago
4 0

Answer:The magnitude of the impulse applied to stop the cart is 50 N sec.

Explanation:

Force applied on the cart = 10 N

Interval of time till force applied ,Time= 5 second

Impulse=Force\times Time=10 N\times 5 seconds= 50 N seconds

The magnitude of the impulse applied to stop the cart is 50 N sec.

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A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

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Q = ba⁴ * ε₀

5 0
2 years ago
A jet engine gets its thrust by taking in air, heating and compressing it, and
nadya68 [22]

Answer:

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

Explanation:

Thrust is known as a reaction force which appears when a system expels or accelerates mass in one specific direction. If we know the acceleration and the mass of the air expelled by the jet engine, we can compute the thrust .

The acceleration is calculated by using the dynamics formula

\displaystyle a=\frac{v_f-v_o}{t}

The values are  

v_f=500\ m/s,\ v_o=100\ m/s,\ t=1\ sec

\displaystyle a=\frac{500-100}{1}=400\ m/s^2

The thrust is

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

4 0
2 years ago
What is the tangential velocity at the edge of a disk of radius 10cm when it spins with a frequency of 10Hz? Give your answer wi
Nina [5.8K]

Answer:

630cm/s

Explanation:

In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr

ὦ is the angular velocity = 2πf

r is the radius of the disk

f is the frequency

Given the radius of disk = 10cm

frequency = 10Hz

v = 2πfr

v = 2π×10×10

v = 200π

v = 628.32 cm/s

The tangential velocity = 630cm/s ( to 2 significant figures)

8 0
2 years ago
It took a squirrel 0.50\,\text s0.50s0, point, 50, start text, s, end text to run 5.0\,\text m5.0m5, point, 0, start text, m, en
STALIN [3.7K]

Answer:

-5.0m/s

Explanation:

3 0
2 years ago
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Mazyrski [523]
A heat pump absorbs heat from the cold outdoors at 3 C and supplies heat to a
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