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3 years ago

an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

2 answers:

7 0

60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N

Ainat [17]3 years ago

6 0

Answer:

The weight on planet x is 60 N

Explanation:

It is given that,

Weight of an object on earth, W = 98 N

Value of acceleration due to gravity, $g=9.8\ m/s^2$

Weight, $W=mg$

$m=\dfrac{W}{g}=\dfrac{98\ N}{9.8\ m/s^2}=10\ kg$

Since, mass of an object remains same everywhere.

If the value of acceleration due to gravity on planet x is, $a=6\ m/s^2$

So, weight on the planet x is,

$W=m\times a$

$W=10\ kg\times 6\ m/s^2$

W = 60 N

So, the weight on planet x is 60 N. Hence, this is the required solution.

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A ball is projected horizontally from the top of a cliff. At the same moment, a second identical ball is dropped from rest from

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Answer:3

Explanation:

First ball is thrown with horizontal velocity while other ball is dropped from cliff such that both have zero vertical velocity. So both balls have to cover a distance equal to the height of cliff with same initial velocity.

time taken is given by $t=\sqrt{\frac{2h}{g}}$

where h=height of cliff

g=acceleration due to gravity

horizontal velocity to first ball will make the ball to travel more horizontal distance as compared to second ball.

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3 years ago

12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L

Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0

3 years ago

An athlete completes 1 laps around a track with a radius of 25 meters in 180 seconds. What is the magnitude of the athlete's tan

DanielleElmas [232]

Answer:

0.872m/s

Explanation:

Tangential velocity is given by the formula,

$v= 2\pi r/ t$

In the question given,

radius= 25meters

time= 180secs

pie= 3.14

number of laps= 1

The magnitude of tangential velocity equals;

$\frac{1lap* 2 *3.14 *25m}{180secs}$

v = 157m/180secs

Therefore, the magnitude of the tangential velocity

=0.872m/secs

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Explanation:

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