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3 years ago

what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2

1 answer:

monitta3 years ago

5 0

On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N

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A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the elli

Volgvan

Answer:

Explanation:

height of Ellipse $=30 cm$

i.e. $2 a=30$

Width of Ellipse $=22 cm$

i.e. $2 b=22$

Equation of a vertical Ellipse is

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

$a=15 ,b =11$

$\frac{x^2}{11^2}+\frac{y^2}{15^2}=1$

at $y=4 cm$

$\frac{x^2}{11^2}+\frac{4^2}{15^2}=1$

$x=\frac{11}{15}\times \sqrt{15^2-4^2}$

$x=10.6 cm$

5 0

2 years ago

How fast can a man run in miles per hour?

ExtremeBDS [4]

Depends on how well built he is probably for the average American 8 MPH

6 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

A train accelerates to a velocity of 500 m/s over time of 2s. The acceleration it experienced was 50m/s2. What was its initial v

ValentinkaMS [17]

Explanation:

We know ,

v = u + at
u = v - at
u = 500 - 2*50
u = 400 m/s

7 0

2 years ago

HURRY PEASE what happens to current when there is a diode on the circuit and why?

kolbaska11 [484]

Answer:

The current can't 'split down the parallel branch, because the diode is reverse biased so is blocking the flow of current. So basically it's acting as an open circuit. Also when the current flows it wouldn't reduce the currents amount flow through the resistor.

Explanation:

6 0

2 years ago