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kozerog [31]
4 years ago
10

An object with a momentum of 1500 kg-m/s directed east is acted upon by an impulse of 100.0 kg-m/s in the same direction. What i

s the final momentum of the object? 14. A 100 kg ball is initially traveling at 10 m/s. If the ball experiences a constant external force that increases the ball’s velocity to 15 m/s, what is the net impulse that acted on the ball?
Physics
1 answer:
fiasKO [112]4 years ago
7 0

Answer:

The final momentum of the body = 1600 kgm/s

Thus the impulse that acted on the body. = 500 N.s

Explanation:

Momentum: This can be defined as the product of mass and velocity. The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Mathematically, momentum can be expressed as

M =mv.

Where M = momentum of the object, m = mass of the object, v = velocity of the object.

Impulse acting on the object = Final momentum of the object - initial momentum of the object

I = M₂ - M₁

M₂ = I + M₁......................... Equation 1

Where I = impulse, M₁ and M₂ = Final and initial momentum.

Note:

(i) The momentum and impulse act in the same direction

(ii) impulse is also a vector quantity.

Given: M₁= 1500 kgm/s, I = 100 kgm/s.

Substituting these values into equation 1

M₂ = 1500 + 100

M₂ = 1600 kgm/s.

Thus the final momentum of the body = 1600 kgm/s

14.

I = m(v-u)............................................... Equation 2.

Where I = impulse on the object, m = mass of the object, v = final velocity, u = initial velocity.

Given: m = 100 kg, v = 15 m/s, u = 10 m/s.

I = 100(15-10)

I = 100(5)

I = 500 N.s or 500 kgm/s.

Thus the impulse that acted on the body. = 500 N.s

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Llana [10]

Answer;

= 4 m is the length of the man's shadow.

Explanation;

2/x=6/(8+x) cross multiply.

6x=2(8+x)

6x=16+2x

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4x=16

x=2=16/4

x=4 m. is the length of the man's shadow.

3 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

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3 years ago
A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
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A=0.05.0M.

B=68.9244GPE.34

C=0

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3 years ago
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
sammy [17]

Answer:

(a) emf_L=-LI_{max}\omega cos(\omega t)

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

I(t)=I_{max}sin(\omega t)   (1)

(a) The induced emf is given by the following formula

emf_L=-L\frac{dI}{dt}    (2)

You derivative the expression (1) in the expression (2):

emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)

(b) At t=0 the current is zero

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w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0
3 years ago
A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?​
Shalnov [3]

Answer:

12.5

Explanation:

given

distance=25

time=2sec

required

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solution

=v=s/t

25/2

=12.5m/s

3 0
3 years ago
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