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Daniel [21]
3 years ago
10

A 250 kg car has 6875 kg•m/s of momentum. What is it’s velocity?

Physics
1 answer:
liraira [26]3 years ago
5 0

Answer:

v = 27.5 m/s

Explanation:

p = m × v

6,875 = 250 × v

250v = 6,875

v = 6,875/250

v = 27.5 m/s

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Please help ASAP
Mandarinka [93]

Answer:

5cm by 4cm by 10cm = 200

200 / 10 = 20

20 :>

6 0
3 years ago
As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o
STatiana [176]

Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2

1\ N/m^2=\frac{1}{6894.757}\ lb/in^2

3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0
3 years ago
Some of the largest volcanoes in the solar system are on mars. The most likely explanation is that mars
ivolga24 [154]

Answer: Mars has lower gravity and lower magma buoyancy

Explanation: The gravity in mars is low. Low gravity affect the occurrence of volcanic eruptions. The buoyancy of the magma in Mars is lower and the depth of the magma chamber deeper (Buoyancy is the difference in density between the surrounding crust and the magma ascending for eruption). Olympus Mons which is larger and wider than mount Everest was formed as a result of overcoming of the low gravity and buoyancy by the magma to release enormous volume of magma to the surface.

Another reason why the volcanoes on Mars are so large is because the crust on Mars is static unlike Earth.

4 0
3 years ago
Two charged particles are accelerated through a uniform electric field and zero magnetic field, then enter a region with zero el
Mrac [35]

Answer:

Explanation:

In magnetic field , charged particle will have circular path . Let the radius of their circular path be r₁ and r₂ . Let their velocity at the time of entering magnetic field be v₁ and v₂ .

The velocity with which they will come out of electric field can be measured from following equation

Eq = 1/2 m v²  , E is electric field , q is charge on the particle , m is mass and v is velocity .

v² = 2Eq / m

radius of circular path can be measured by the following expression

m v² / r = Bqv

2Eq / r = Bqv

r = 2Eq / Bqv

= 2E / Bv

r² = 4E² / B²v²

= 4E²m / B²x 2Eq

since E , B and q are constant

r² = K . m

r₂² / r₁² = m₂ / m₁

1.5²

m₂ / m₁ = 1.5²

= 2.25

6 0
3 years ago
three masses are connected by a light string that passes over a frictionless pulley as shown. (a) what is there acceleration of
Ket [755]

The mass on the left has a downslope weight of  

W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N  

The mass on the right has a downslope weight of  

W2 = 8kg * 9.8m/s² * sin35º = 45.0 N  

The net is 25.3 N pulling downslope to the right.  

(a) Therefore we need 25.3 N of friction force.  

Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º  

25.3N = µ * 92.3 N  

µ = 0.274  

(b) total mass is 11.5 kg, and the net force is 25.3 N, so  

acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s²  

tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N  

Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √

hope this helps.   :)

3 0
4 years ago
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