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Tanzania [10]
3 years ago
11

PLZZZ HELP

Physics
1 answer:
Doss [256]3 years ago
4 0

Answer/Explanation:

The weight of an object is defined as the force that is exerted due to the gravitational force.

Mathematically, it can be written as :

W = m g

Where

m is the mass of the object

g is the acceleration due to gravity

Also,  

We know that the value of g varies with respect to the location. At the equator, the value of g is less as compared to the poles.

The feature of an object that affects its weight are :

Mass of the object

Location of the object

How much force Earth exerts on the object

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A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin
Greeley [361]

Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Electric Resistance</u>

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}

This sum is repeated 10 times. Operating each term:

\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}

All the terms have the same denominator, thus:

\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}

Taking the reciprocals:

R_e=R/100

The equivalent resistance of the combination is R/100

6 0
3 years ago
How could you determine the instantaneous speed during a<br> trip?
Fiesta28 [93]
Look at your speedometer for say, a couple of seconds. Depends on whether or not you are moving on average at a constant speed (speedo won't change much) or whether you're in a polluting traffic jam/queue in which case the speedo will go up and down like a yo yo. to determine the speed, you'd probably need to plot the speed on the speedo against the times at which the speedo speeds were read from the speedo.
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There's more to motion than simply changing position. True Or False
prohojiy [21]

Answer: I think that it is False, if its wrong I am sorry.

Explanation:

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What happens when you brake earphones in to smaller and smaller pieces
nikklg [1K]
They dont work because you crushed them duh
5 0
3 years ago
A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu
Katyanochek1 [597]

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

                                   V_f = sqrt(650.9861)

                                  V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0
3 years ago
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