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2 years ago

What is the instantaneous velocity v of the particle at t=10.0s?

1 answer:

8 0

Instantaneous velocity is defined at a specific value of time t. It can never be observed or measure. To know a velocity, one must know its distance and velocity. It is represented by the equation velocity is equal to delta distance over delta time. The problem above lacks data.

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A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?

lana [24]

KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps

7 0

2 years ago

Read 2 more answers

If you have to apply 40n of force on a crowbar to lift a rock that weights 400n, what is the actual mechanical advantage of the

Mrrafil [7]

The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
$MA= \frac{F_{out}}{F_{in}}$
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
$MA= \frac{400 N}{40 N}=10$

3 0

2 years ago

An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th

maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

3 0

2 years ago

A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the

Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0

1 year ago

A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char

Aleksandr [31]

Answer:

Ek = 8,79 [J]

Explanation:

We are going to solve this problem, using the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁ = Ek + U₁

As charge are at rest Ek = 0

And U₁ has two components

U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then E₁ = E₁₂ + E₃₂

E₁ = 10,5 [J]

At the moment of Q₂ passing x = 40 cm or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂ = U₁₂ + U₃₂

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3

U₃₂ = - 9*20*10⁻²/0,3

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as

E₂ = Ek + U₂ and E₂ = E₁

Then

Ek + U₂ = E₁

Ek = 10,5 - U₂

Ek = 10,5 - 1,71

Ek = 8,79 [J]

5 0

1 year ago