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2 years ago

What observations tell you that a car is accelerating

Physics

1 answer:

professor190 [17]2 years ago

8 0

When the car moves and makes a sound that is louder that when the car is just sitting there

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By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)

Fofino [41]

Gravity on the surface of sun is given as

$g = \frac{GM}{R^2}$

here we know that

$M = 1.98 \times 10^{30} kg$

$R = 6.95 \times 10^8 m$

now we will have

$g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}$

$g = 273.4 m/s^2$

now we need to find the ratio of weight on surface of sun and on surface of Earth

$R = \frac{mg_{sun}}{mg_{earth}}$

$R = \frac{273.4}{9.8} = 28 times$

so weight will increase by 28 times

6 0

2 years ago

Please help for 21! will mark brainliest

Katyanochek1 [597]

The answer for this question is D

5 0

2 years ago

What do you know about a laser to predict its performance as a surgical tool? Give at least four factors and explain the signifi

Akimi4 [234]

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5 0

2 years ago

An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in

weqwewe [10]

Answer:

a) $t=1.4\times 10^{-5}\ s$

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

$t=1.4\times 10^{-5}\ s$

We know that

F = m a = E q ------------1

Mass of electron ,m

$m=9.1\times 10^{-31}\ kg$

Charge on electron

$q=1.6\times 10^{-19}\ C$

So now by putting the values in equation 1

$a=\dfrac{E q}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 0.027}{9.1\times 10^{-31}}\ m/s^2$

$a=4.74\times 10^{9}\ m/s^2$

$S= ut+\dfrac{1}{2}at^2$

Here initial velocity u= 0 m/s

$S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m$

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0

3 years ago

Read 2 more answers

A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e

jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?$

Putting values in above equation, we get:

$\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K$

Hence, the final temperature is 350 K

8 0

2 years ago