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2 years ago

A window washer who does not want to change his position will want the forces acting on him to be ____________.

Physics

1 answer:

natali 33 [55]2 years ago

6 0

My answer is a balanced

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A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebou

Vsevolod [243]

Answer:

$\Delta p=2.27\frac{kg\cdot m}{s}$

Explanation:

The momentum change is defined as:

$\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)(1)$

Taking the downward motion as negative and the upward motion as positive, we have:

$v_f=2\frac{m}{s}(2)\\v_i=-2\frac{m}{s}(3)$

Replacing (2) and (3) in (1):

$\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}$

5 0

3 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the

vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁ $ln[\frac{P_{1} }{P_{2} }]$

= mRT₁ $ln[\frac{P_{1} }{P_{2} }]$ [∵pV = mRT]

= (5) (0.287) (272.039) $ln[\frac{2.0}{1.0}]$

= 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU = $mc_{v}(T_{2} - T_{1} ) = 0$

From 1st law of thermodynamics

Q = ΔU + W

= 0 + 270.588

= 270.588 kJ

4 0

3 years ago

In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

$\Delta U=q+w\\\\U_{final}-U_{initial}=q+w$

where,

$U_{final}$ = final internal energy

$U_{initial}$ = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and $U_{initial}$

initial internal energy = $U_{initial}$ = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

$U_{final}-(2000J)=(1000J)+(500J)$

(4) By solving the terms, we get

$U_{final}-(2000J)=(1000J)+(500J)$

$U_{final}-(2000J)=1500J$

$U_{final}=2000J+1500J$

$U_{final}=3500J$

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0

3 years ago

Is the star moving toward earth, away from earth, or is there not enough information provided to determine its motion?

ivanzaharov [21]

If a star is moving towards Earth, shift towards the blue end of the spectrum, this is called blue shift. If the star is moving away from Earth the light from that star will be red and is called red shift .

The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum

if a star is moving towards Earth, it appears to emit light that is shorter in wavelength compared to a source of light that isn't moving. Because shorter wavelengths correspond to a shift towards the blue end of the spectrum, this is called blue shift.

If the star is moving away from Earth, its light will lose energy to reach Earth, therefore the light from that star will be red and is called red shift

learn more about blue shift :

brainly.com/question/5368237?referrer=searchResults

#SPJ4

8 0

1 year ago

A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N.

balandron [24]

Complete Question:

A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N. (8 marks)

a) What is the downward force of gravity on the rocket at blast-off?

b) What is the unbalanced force on the rocket at blast-off?

c) What is the acceleration of the rocket as it leaves the launching pad?

d) As the rocket travels upwards, the engine thrust remains constant, but the mass of the rocket decreases. Why?

e) Does the acceleration of the rocket increase, decrease, or remain the same as the engines continue to fire?

Answer:

a) $-2.8616 \times 10^{7} N$ is the downward force of gravity on the rocket at blast-off.

b) $4.784 \times 10^{6} N$ is the unbalanced force on the rocket at blast-off

c) $1.638 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the rocket as it leaves the launching pad

d) Because the propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases when the rocket moves upward.

e) The acceleration of the rocket increases when engines continue to fire

Explanation:

Given:

Mass (m) = $2.92 \times 10^{6} \mathrm{kg}$

a) In physics, weight can be defined as the applied force on a body by gravity. It is the product of mass (m) and gravity $\left(g=9.8 \mathrm{m} / \mathrm{s}^{2}\right)$

$\text { weight }(W)=m \times g=2.92 \times 10^{6} \times(-9.8)=28.616 \times 10^{6}=-2.8616 \times 10^{7} N$

The negative sign indicates the downward force of gravity.

b) To find the unbalanced force on the rocket at blast-off,

Accelerating force,

$F_{a}=F+W=3.34 \times 10^{7}+\left(-2.8616 \times 10^{7}\right)=(3.34-2.8616) \times 10^{7}$

$F_{a}=0.4784 \times 10^{7}=4.784 \times 10^{6} N$

c) Newton’s second law of motion states that the object’s acceleration depends on two variable:

Directly proportionate to the object’s force existed
Inversely proportionate to the mass of the objects

The equation can be given as below,

$Force =m \times acceleration$

$\text { Acceleration }=\frac{F_{a}}{m}=\frac{4.784 \times 10^{6}}{2.92 \times 10^{6}}=1.638 \mathrm{m} / \mathrm{s}^{2}$

d) The pushing of rocket upward will happen as long as the engine gets fired. The propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases (taotal mass) when the rocket moves upward.

e) The acceleration of the rocket increases when engines continue to fire

Let consider $F_{a}$ is constant, mass gets decreasing, then the acceleration would be increasing (as mass and acceleration are inversely proportionate to each other) .

8 0

3 years ago