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zmey [24]
2 years ago
12

A force of constant magnitude F and fixed direction acts on an object of mass m that is initially at rest. If the force acts for

a time interval ∆t over a displacement ∆x , what is the magnitude of the resultant change in the linear momentum of the object?
Physics
1 answer:
ZanzabumX [31]2 years ago
5 0

Answer:

\Delta P = F \Delta t

Explanation:

As we know Newton's II law

F= Rate of change in momentum

so we will have

F = \frac{\Delta P}{\Delta t}

now we will have

\Delta P = F \Delta t

so here we can say that change in momentum of the object is the product of force and interval of time for which the force is acting on it.

so we will have

\Delta P = F \Delta t

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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0
2 years ago
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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.
Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

\sin\theta=\dfrac{v'}{v}

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

\sin\theta=\dfrac{75}{310}

\theta=\sin^{-1}(\dfrac{75}{310})

\theta=14.0^{\circ}

We need to calculate the resultant speed

Using formula of resultant speed

\cos\theta=\dfrac{v''}{v}

Put the value into the formula

\cos14=\dfrac{v''}{310}

v''=\cos14\times310

v''=300.79\ km/h

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0
2 years ago
A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow
Ymorist [56]

Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

W is the weight of the crate

As P is unable to move the Block therefore Normal reaction keeps on acting on block

thus we can say that

P-W+C=0

P=W-C

                   

7 0
2 years ago
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In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Ugo [173]

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad  = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[  9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad  = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0
2 years ago
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

mgh=\dfrac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 35.4}

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
2 years ago
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