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4 years ago

If you are given force and time, you can determine power if you can know...

Physics

2 answers:

jek_recluse [69]4 years ago

8 0

The answer would be watts

i hope this helps :))

Vera_Pavlovna [14]4 years ago

6 0

If you're given the force and the length of time that the force acted,
then you can determine momentum, with units of [mass-length/time].

Without more information, you can't determine power, watts, energy,
joules, or distance.

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Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

4 0

4 years ago

Create a model in the space below that demonstrates how action potentials ensue. A sample model can be found under the Unit Pack

Pepsi [2]

Answer:

K+NA+30

Explanation:

8 0

3 years ago

What is the difference between transverse and longitudinal waves?

Fed [463]

To explain how transverse and longitudinal waves work, let us give two examples for each particular case.

In the case of transverse waves, the displacement of the medium is PERPENDICULAR to the direction of the wave. One way to visualize this effect is when you have a rope and between two people the rope is shaken horizontally. The shift is done from top to bottom. This phenomenon is common to see it in solids but rarely in liquids and gases. A common application usually occurs in electromagnetic radiation.

On the other hand in the longitudinal waves the displacement of the medium is PARALLEL to the direction of propagation of the wave. A clear example of this phenomenon is when a Slinky is pushed along a table where each of the rings will also move. From practice, sound waves enclose the definition of longitudinal wave displacement.

Therefore the correct answer is:

C. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.

8 0

4 years ago

Let σa, σb, σc, and σd be the respective surface charge densities on surfaces A, B, C, and D. Take all the charge densities to b

aniked [119]

Answer:

a) σa−σb−σc−σd=0

Explanation:

The parallel plate capacitor is the one in which two metal plates are connected in parallel with some distancing among them. The electric field from both plates is denoted by E = σ / 2ϵ0. The σ is the charge density. The Electric field in plate I will vanish when the surface charge of σa is positive and rest of the charges are negative. The correct option is a.

7 0

3 years ago

A student removes a 10.5 kg stereo amplifier from a shelf that is 1.82 m high. The amplifier is lowered at a constant speed to a

Nonamiya [84]

Answer:

(a) the work done by the student is 110.1 J

(b) The gravitational force that acts on the amplifier is 102.9 N

Explanation:

Given;

mass of the amplifier, m = 10.5 kg

initial position of the amplifier, x₀ = 1.82 m

final position of the amplifier, x₁ =0.75 m

The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m

(b) The gravitational force that acts on the amplifier;

F = mg

F = 10.5 x 9.8

F = 102.9 N

(a) the work done by the student is calculated as;

W = FΔx

W = 102.9 x 1.07

W = 110.1 J

7 0

3 years ago