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dexar [7]
3 years ago
5

food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;

about 1/6 of the value on earth. Compute thefollowing:a.The force exerted by the product on the earth’s surface; inSI units and English units.b.The product force exerted on the surface of the moon; inSI and English units
Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N

Converting to lbf

98.1\times 0.22481=22.053861\ lbf

On Moon

W=10\times 1.624\\\Rightarrow W=16.24\ N

Converting to lbf

16.24\times 0.22481=3.6509144\ lbf

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

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C. Magnetism

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3 years ago
How much will the kinetic energy of a school bus increase if its velocity is tripled?
son4ous [18]

Answer:

Explanation:

Given a school bus.

Let say initially the school bus is traveling with speed "v"

Let assume mass of school bus is "m"

Then, the initial kinetic energy is

K.E_initial = ½mv²

Now, if the initial velocity is tripled,

Then, the new velocity is

v_new = 3v.

Note: the mass of the school does not change it is constant

Then, new kinetic energy is

K.E_new = ½m(v_new)²

v_new = 3v

Then,

K.E_new = ½m(3v)²

K.E_new = ½m × 9v²

K.E_new = 9 × ½mv²

Since K.E = ½mv²

Then,

K.E_new = 9 × K.E

So, the new kinetic energy will be 9 times the initial kinetic energy.

So, option D is correct

D. It will be nine times greater.

4 0
3 years ago
Is a measure of waves that passes a point in a given amount of time
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8 0
3 years ago
Projectile Motion: A hobbyist launches a projectile from ground level on a horizontal plain. It reaches a maximum height of 72.3
marin [14]

Answer:

The angle of launch from the horizontal direction is 20.99° .

Explanation:

Let u and θ be the initial speed and angle of projection from the horizontal axis of the object respectively.

The equations for projectile motion are :

H = ( u² sin²θ)/ 2g      ......(1)

Here H is maximum height of the projectile motion and g is acceleration due to gravity.

R = ( u² sin2θ)/g         .......(2)

Here R is the maximum horizontal displacement of the object.

Rearrange equation (1) in terms of u².

u² = (2gH)/sin²θ

Substitute this equation in equation (2).

R = (2gH sin2θ) / (sin²θ x g)

R = (2H sin2θ)/sin²θ

Using trigonometry property, sin2θ = 2 cosθ sinθ

So, above equation becomes,

R = (2H x 2 cosθ sinθ)/sin²θ

R = (4H cosθ)/sinθ

tanθ = R/4H

θ = tan⁻¹(R/4H)

Substitute 111 m for R and 72.3 m for H in the above equation.

θ = tan⁻¹( 111/ 4 x 72.3 )

θ = tan⁻¹(0.38)

θ = 20.99°

3 0
3 years ago
An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t
lana [24]

Answer:

Equal to the weight of the object

Explanation:

There are two forces acting on the object hanging on the string:

- The tension in the string, T, acting upward

- The weight of the object, mg, acting downward

We can write Newton's second law as:

T-mg=ma

where a is the acceleration of the object.

Here we are told that the elevator is moving at constant speed: this means that the acceleration of the object is zero, therefore a = 0 and the equation becomes

T=mg

therefore, the tension in the string is equal to the weight of the object.

7 0
3 years ago
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