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3 years ago

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Steven carefully places a m = 1.85 kg wooden block on a frictionless ramp so that the block begins to slide down the ramp from r

est. The ramp makes an angle of θ = 51.3 ° up from the horizontal. Which forces do nonzero work on the block as it slides down the ramp?

1 answer:

Helen [10]3 years ago

4 0

Answer:

Gravity force

Explanation:

We are given that

Mass of wooden block,m=1.85 kg

$\theta=51.3^{\circ}$

We have to find the force which do non zero work on the block as it slides down the ramp.

When a block slides down the ramp on a frictionless surface .

Then, weight act on the wooden block in downward direction.

We know that weight is that force which act on the body due to gravity.

Then, we can say that the force which do nonzero work on the block as it slides down the ramp is gravity force.

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A helicopter is ascending vertically. a passenger accidentally drops her wallet out the sides of the helicopter when it is 160 m

Advocard [28]

Answer:

(E)56.0 m/s

Explanation:

Height =h=-160 m

Because the wallet moving in downward direction

Time=t=7 s

Final speed of wallet=v=0

We have to find the speed of helicopter ascending at the moment when the passenger let go of the wallet.

$v^2-u^2=2gh$

Where $g=9.8 m/s^2$

Substitute the values

$0-u^2=2(-160)\times 9.8$

$u^2=3136$

$u=\sqrt{3136}=56m/s$

Option (E) is true

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3 years ago

A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is

snow_lady [41]

Answer:

$\bf\pink{10\:m/s}$

Explanation:

<h2><u>Given</u> :-</h2>

$\sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}$
$\sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}$

<h2><u>To Find</u> :-</h2>

$\sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}$

<h2><u>Formula to be used</u> :-</h2>

$\sf\blue{K.E. = \dfrac{1}{2} mv^{2}}$

Where,

K.E. = Kinetic energy possessed by the body
M = Mass of the body
V = Velocity of the body

<h2><u>Solution</u> :-</h2>

$\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}$

$\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}$

$\to\:\:\sf\green{5000 = 50 \times v^{2}}$

$\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}$

$\to\:\:\sf\purple{100 = v^{2}}$

$\to\:\:\sf\red{\sqrt{100} = v}$

$\to\:\:\sf\orange{ 10 = v}$

$\to\:\:\bf\pink{v = 10\:m/s}$

Velocity of the vehicle at the instant is $\bf\green{10\:m/s}$

7 0

2 years ago

Read 2 more answers

A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th

defon

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

8 0

3 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

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3 years ago

On a smooth horizontal floor, an object slides into a spring which is attached to another mass that is initially stationary. Whe

HACTEHA [7]

Answer:

E) momentum and mechanical energy

Explanation:

In the context, an object is attached to the another mass with a spring which is initially at a rest position. Now when the spring is compressed, the two masses moves with the same speed. Now since the both the masses combines with the spring to move together they are considered as one system and in this case the momentum and the kinetic energy will be conserved.

The kinetic energy and momentum of the system after collision and the kinetic energy and momentum of the two masses before collision will be constant.

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3 years ago