Sentence using lunar for a kid

The mass of an object is 60kg on the surface of the earth what will be its weight on the surface of the moon

iris [78.8K]

Answer:

Wm = 97.2 [N]

Explanation:

We must make it clear that mass and weight are two different terms, the mass is always preserved that is to say this will never vary regardless of the location of the object. While weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 10 [m/s²]

But in order to calculate the weight of the body on the moon, we must know the gravitational acceleration of the moon. Performing a search of this value on the internet, we find that the moon's gravity is.

gm = 1.62 [m/s²]

Wm = 60*1.62

Wm = 97.2 [N]

8 0

3 years ago

Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc

kaheart [24]

Answer:

The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.

(3) and (4) is correct option.

Explanation:

Given that,

Beat frequency f = 5.0 Hz

Frequency f'= 462 Hz

We need to calculate the possible frequencies for the A string of the other violinist

Using formula of frequency

$f'=f_{1}-f$ ...(I)

$f'=f_{1}+f$ ...(II)

Where, f= beat frequency

f₁ = frequency

Put the value in both equations

$f'=462-5=457\ Hz$

$f'=462+5=467\ Hz$

Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.

4 0

3 years ago

In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas

Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0

3 years ago

Read 2 more answers

A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t

weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

$\theta=90^{\circ}$

We have to find the their velocity immediately after the tackle.

Initial momentum, $P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s$

According to law of conservation of momentum

Initial momentum=Final momentum= $(m+m')V$

$627.6=(85+90)V=175V$

$V=\frac{627.6}{175}=3.59 m/s$

3 0

3 years ago

A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.

ycow [4]

Answer:

σ = 3.402 KPa , γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

Find the shearing stress, shearing strain and shear modulus.

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

- The shear effect results in a stress in the gelatin block.

- The formulation of stress ( σ ) is given below:

σ = F / A

Where,

A : The surface area of the object that experiences the shear force.

- The top surface have the following dimensions:

A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

σ = 49 / 0.0144

σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

γ = x / L

γ = 10 / 40

γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- It is mathematically expressed as a ratio of shear stress ( σ ) and shear strain ( γ ):

G = σ / γ

G = 3.402 / 0.25

G = 13.608 KPa

7 0

3 years ago