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3 years ago

A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

Physics

1 answer:

joja [24]3 years ago

6 0

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

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You are trying to overhear a juicy conversation, but from your distance of 25.0 m , it sounds like only an average whisper of 20

spayn [35]

Answer:So You Decide To Move Closer To Give The Conversation A Sound Level Of 80.0dB Instead. ... You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.0dB .

Explanation:

7 0

3 years ago

What is the potential energy of an object 20 m in the air with a<br> mass of 600 kg?

Lana71 [14]

Answer:

Ep = 117600 J

Explanation:

Data:

Mass (m) = 600 kg
Height (h) = 20 m
Gravity (g) = 9.8 m/s²
Potential Energy (Ep) = ?

Use formula:

Ep = m * g * h

Replace:

Ep = 600 kg * 9.8 m/s² * 20 m

Multiply operations, and units:

Ep = 117600 J

What is the potential energy?

The potential energy is <u>117600 Joules.</u>

7 0

2 years ago

Select the correct answer.

Simora [160]

Answer:

A. 2.36 Newtons

Explanation:

F = GmM/d²

F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²

F = 2.36121...

Very poor question design.

mass of box... 1 significant digit

distance... 2 significant digits

mass of earth... 3 significant digits

value of G... 4 significant digits

Answer precision to 3 significant digits is not justifiable

7 0

3 years ago

Density: Two blocks, A and B, are put in a tank of water. Block A has a density of 1.21 g/cm³. Block B has a density of 1.37 g/c

stira [4]

Answer:

Block A

Explanation:

Block A will float higher in the water compared to the second Block.

The density of water is 1g/cm³.

According to the principle of floatation "an object that floats in a liquid will displace equal amount of fluid to the weight of the object".

A body will become more submerged in water if it has more density because density is the mass per volume of body.

An object with a higher density than another will sink in the liquid of the one with lesser density.

Object A has lesser density and will float higher up and displace very little water.
Object B has higher density and will be more submerged.

4 0

3 years ago

if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0

3 years ago