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Cerrena [4.2K]
3 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]3 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]3 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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3 years ago
Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A p
serg [7]

Answer:

f = 735 Hz

Explanation:

given,

Person distance from speakers

r₁ = 4.1 m      r₂ = 4.8 m

Path difference

d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m

For destructive interference

d = \dfrac{n\lambda}{2}

where, n = 1, 3,5..

we know, λ = v/f

d = \dfrac{n v}{2f}

v is the speed of the sound = 343 m/s

f is the frequency

f = \dfrac{n v}{2d}

for n = 1

f = \dfrac{343}{2\times 0.7}

     f = 245 Hz

for n = 3

f = \dfrac{3\times 343}{2\times 0.7}

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Hence,the second lowest frequency of the destructive interference is 735 Hz.

7 0
4 years ago
Why is Radiology important in the medical field?
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5 0
2 years ago
A .005kg projectile leaves a 1500kg launcher with a velocity of 750 m/s. What is the recoil velocity of the projectile
docker41 [41]

Answer:

The recoil velocity of the projectile is 0.0025m/s

Explanation:

Given:

Mass of the projectile =0.005kg

Mass of the launcher = 1500kg

Velocity =  750 m/s.

To Find:

The recoil velocity of the projectile = ?

Solution:

The recoil velocity is the obtained by dividing the "recoil momentum"  by the "mass of the recoil body".  The recoil momentum is equal to the momentum of the other body. The momentum of the other body is equal to it mass times its velocity.

Lets find the recoil momentum,

Recoil momentum = mass of the projectile X velocity

Recoil momentum =0.005 \times 750

Recoil momentum = 3.75

Now Recoil Velocity,

Recoil Velocity = \frac{\text { Recoil Momentum}}{\text {Mass of the launcher}}

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Recoil Velocity = 0.0025m/s

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4 years ago
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