A person burns 2000 J of energy running 50 meters in 17 seconds at a constant velocity. What was the average force they applied?

27. Improvements in space technology have greatly helped scientists better understand the stars, planets, and other

stellarik [79]

Answer:

B

Explanation:

Technological advancements in astronomy have led to the development of thousands of products such as

new materials, medical devices, and communications satellites.

For example, the production of instruments like telescopes, microscopes, and other measuring devices now enables us to identify things that we couldn't examine with the naked eye.

Similarly, the development of communication satellites has led to the emergence and improvements in the mobile network industry.

8 0

3 years ago

For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul

loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

$\frac{P_{o} }{P_{1} } =36.73$ (M₁ = 3)

If Po₁ = Po₂

$\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825$

Table A-1:

$\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63$

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0

3 years ago

A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?

alexandr1967 [171]

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

$v=u+at$

Here, train starts from rest so, u = 0

$v=0+1.5\ m/s^2\times 5\ s$

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

3 years ago

Match each situation

EastWind [94]

A: is potential
C: is losing kinetic energy and gaining potential energy
B: kinetic energy is at its highest
D: is loosing potential energy and gaining kinetic energy

3 0

3 years ago