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Leya [2.2K]
3 years ago
8

Your car gets a flat! You go from 90 kilometers per hour to a stop in 6 seconds. What is your rate of deceleration?

Physics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

The rate of deceleration is 15 km/h^2.

Explanation:

Since the car comes in 90km/h, it will need deceleration for 15 km/h for 6 seconds to finally stop the car. 15*6=90

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The protons in a nucleus are approximately 2 ✕ 10^−15 m apart. Consider the case where the protons are a distance d = 1.93 ✕ 10^
Bond [772]

Answer: 61.2N

F=kq1q2/r^2

F=8.98E9(1.6E-19)(1.6E-19)/(1.93E-15)^2

F=61.2N

6 0
3 years ago
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Determine the gravitational force of attraction between the Earth and the Sun given the mass of the earth is 6 x 10^24 kg, the m
Helga [31]

Answer:

35.57*10^21 N.

Explanation:

F = Gmm/r², where G is 6.67*10e-11 & r is distance.

F = 6.67*10^-11 * 6*10^24 *2*10^30 /(1.5*10^11)².

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Which is the correct formula for finding the frequency of an electromagnetic wave? f = c – f = f = + c f =
liubo4ka [24]
The answer for the online test is going to be the last one, D
6 0
3 years ago
Gold has a specific heat of 0.130 J/g*C. If 195 joules of heat are added to 15 grams of gold how much does the temperature of th
Anna71 [15]

The correct answer to the question is :  100\ ^0C

EXPLANATION :

As per the question, the specific heat of gold is given as c = 0.130\ J/g^0C

The heat given to the gold dQ = 195 J

The mass of the gold is given as m = 15 gram.

We are asked to calculate the change in temperature.

Let the change in temperature is dT.

We know that dQ = mcdT

                      dT=\ \frac{dQ}{mc}

                             =\ \frac{195}{15\times 0.130}

                             =\ 100\ ^0C                   [ANS]

Hence, the change in temperature is 100 degree celsius.

5 0
3 years ago
If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency
Pepsi [2]

Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = - \frac{ d \Phi_B}{dt}

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

6 0
3 years ago
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