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3 years ago

The momentum of a 3000 kg truck is 6.36 x 104 kg·m/s. At what speed is the truck traveling? m/s

Physics

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

21.2m/s

Explanation

Given data

P=6.36 x 10⁴kg·m/s.

M=3000kg

From the momentum expression

P=mv

v=p/m

Substitute

v=63600/3000

v=21.2m/s

Hence the Velocity is 21.2m/s

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A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"

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Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

$mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s$

The velocity of the coaster is 0.29713 m/s

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3 years ago

A cloud of interstellar gas is rotating. Because the gravitational force pulls the gas particles together, the cloud shrinks, an

Fynjy0 [20]

Answer:

Greater than

Explanation:

Here, angular momentum is conserved.

$l_1\omega_1 =l_2\omega_2$

When the cloud shrinks under the right conditions, a star may be formed.

Thus, Diameter of clouds are much higher than a star.

Moment of inertia of cloud is greater than the star's inertial.

so, angular velocity of the star would be greater than angular velocity of the rotating gas.

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3 years ago

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A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

anzhelika [568]

Answer:

$\phi_i = BA$

Explanation:

magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

$\phi = \vec B. \vec A$

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

$\phi = BAcos0 = BA$

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3 years ago

the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su

CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ = 1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

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