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professor190 [17]
3 years ago
12

A race car starts from rest and accelerates uniformly to 69 mph in 4.5 s.

Physics
1 answer:
Marizza181 [45]3 years ago
5 0
For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2

If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2
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True or False – All transformations fit the Law of Conservation of Energy.
Nataly [62]

Answer:

The correct answer is true.

5 0
3 years ago
Describe how an oscilloscope should be used to measure the frequency of the sound wave from the sonometer
ratelena [41]

Answer:

            T = reading (cm) time base (s / cm)

            f = 1 / T

Explanation:

An oscilloscope is a piece of equipment that allows you to visualize and measure a wave that reaches you, in the case of having a sonometer this transforms the sound wave into an electrical signal to be introduced through one of the voltage channels of the equipment, on the screen we will see the oscillating alternating signal, if it is fixed we can make the reading, if it is moving the time base and the trigger must be adjusted to stop it.

In the oscilloscope we can read the period of the signal, this is the time it takes for the signal to repeat itself with this value, we can calculate the frequency with the formula, for the reading of the period the distance is measured on the labeled screen and multiplied by the time base

         

            T = reading (cm) time base (s / cm)

            f = 1 / T

6 0
3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a
Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy  = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m²  ------------------------- (2)

from (1) and (2)

B²q²R² / m²  = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0
3 years ago
Physics 1
Degger [83]

The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.

<h3>What is induced emf?</h3>

Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.

When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.

The given data in the problem is;

B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T

V(velocity)=125 M/SEC

L(length)=25 cm=0.25 m

The maximum emf is found as;

E=VBLsin90°

E=125 × 5.0 × 10⁻⁵ ×0.25

E=1.56 ×10 ⁻³ V

Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V

To learn more about the induced emf, refer to the link;

brainly.com/question/16764848

#SPJ1

7 0
2 years ago
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