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3 years ago

A backpack weighs 8.2 newtons and has a mass of 5kg on the moon. What is the strength of the gravity on the moon?

1 answer:

7 0

Weight is equivalent to the product of the mass of an object and the strength of the gravitational field.

Using:
F = ma

a = 8.2 / 5
a = 1.64 N/kg

The gravitational field strength is equivalent to 1.64 N/kg.

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The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single

Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W = mg = (689/1000)9.8 = 6.75 N, D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0

2 years ago

the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe

Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

β = 10 log $\frac{I}{I_{o}}$

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

I / I₀ = 10 (β / 10)

I = I₀ 10 (β / 10)

trumpet

I1 = 1 10⁻¹² 10 (94/10)

I1 = 2.51 10⁻³ / cm²

Thrombus

I2 = 1 10⁻¹² 10 (107/10)

I2 = 5.01 10-2 W / cm²

low

I3 =1 1-12 (113/10) W/cm²

I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

I_total = I₁ + I₂ + I₃

I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1

I_total = 0.00251 + 0.0501 + 0.1995

I_total = 0.25211 W / cm²

let's bring this amount to the SI system

β = 10 log (0.25211 / 1 10⁻¹²)

β = 114 db

7 0

3 years ago

A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

$M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

Since initial velocity for M1 ball is zero, then

$M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

and

$M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

So, on solving all the above equation, we get an equation for velocity and that is

$\frac{2M_{2}u_{2} }{(M_{1}+M_{2} }$ =final velocity of ball with mass 2.5 kg

$v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s$

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is $\frac{1}{2}*2.5*(4.67)^{2} =27.09 J$

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0

3 years ago

The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat

german

Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of $CsCl$ :

(1) Conversion of solid calcium into gaseous cesium atoms.

$Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)$

$\Delta H_s$ = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

$Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)$

$\Delta H_I$ = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

$Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)$

$\Delta H_D$ = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

$Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)$

$\Delta H_E$ = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

$Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)$

$\Delta H_L$ = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L$

$\Delta H_L=-667KJ/mole$

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

5 0

3 years ago

If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be? A. 200 N forward B.

Korolek [52]

The ball should put 200 N of force towards the golfer.

Newton's Third Law is every action has an equal and opposite reaction.

It's the ball exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

8 0

3 years ago

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