All categories

2 years ago

In the absence of air resistance, ___ accelerates all objects at the same rate of 9.8 m/s2

Physics

1 answer:

jarptica [38.1K]2 years ago

6 0

Answer:

Gravity...this was proven by NASA and was examined by Leonardo da Vinci.

You might be interested in

1. A 9000 kg van was stopped at a traffic light when it rear-ended with an 850 compact car moving to the east at a velocity of 5

____ [38]

Answer:

1.785 m/s

Explanation:

The momentum can be calculated using the expression below

M1 *V1 + M2 * V2 = (M1+M2) V3

M1= mass of van=9000 kg

M2= mass of car= 850kg

V3= velocity of entangled car

V1= Velocity of the van= 0

V2= velocity of the car= 5 m/ s

Substitute the values

(900×0) + (500×5)=( 900+500)× V3

2500=1400 V3

V3=2500/1400

V3= 1.785 m/s

Hence, velocity of the entangled cars after collision is 1.785 m/s

8 0

2 years ago

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will

alukav5142 [94]

Answer:

The change in momentum is $\Delta p = kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A =$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = * (22.9 - 22.6)$

$\Delta p = * (0.3)$

$\Delta p = kg \cdot m/s$

6 0

2 years ago

Which of the following is a risk associated with taking performance-enchancing drugs.

Damm [24]

Answer:

The correct answer is All of the above.

Explanation:

7 0

2 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

5 0

2 years ago

2O POINTS!!! Easy Question. Will Mark Brainiest

wel

Answer: The Electrostatic force of attraction or repulsion between two charges shows that the Newton's third law applies to electrostatic forces.

Explanation: Consider two Oppositely charged charges separated by distance d.

The electrostatic force exerted by charge 1 on charge 2 is.

By Coulomb's Law :

F1 = k $\frac{Q1 Q2}{d2}$ .....................................(1)

The electrostatic force exerted by charge 2 on charge 1 is.

F2 = - k $\frac{Q1 Q2}{d2}$ ................................. (2)

negative sign shows that force are in opposite direction.

From Equation 1 and 2

F1 = - F2

Which implies Newton Third law.

6 0

2 years ago