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Salsk061 [2.6K]
3 years ago
10

If an astronaut travels to different planets, which of the following planets will the astronaut’s weight be the same as on Earth

Physics
2 answers:
AysviL [449]3 years ago
5 0

Answer:

D. none

Explanation:

Nataly_w [17]3 years ago
3 0
<h2>Astronaut travels to different planets - Option 4 </h2>

If an astronaut travels to different planets, none of the planets will the astronaut’s weight be the same as on Earth. On jupiter, weight will be more than the weight on earth. For instance if an astronaut has 100kg on earth then he will have 252 kg on jupiter.

On Mars, weight will be less than the weight on the earth. For instance, if an astronaut has 68 kg on earth then he will has 26 kg on mars. On Mercury, weight of an astronaut will be less than the weight on earth. Example if he has 68 kg on earth then he will have 25.7kg on mercury.

Hence, none of these planets the weight of astronaut will be same as on earth.

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A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg
ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

p=m/V\\\\ p=m/(L*A)\\p*A=m/L

Now the equation for speed of traverse wave is calculated through:

\sqrt \frac{F*L}{m}\\

=\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}

Substituting the values

\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3}  \\

=22.49 m/s

4 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
in a circuit, a 10 resistor is connected in series to a parallel group containing a 60 resistor and a 5 resistor. what is the to
natka813 [3]
The 60 and the 5 in parallel have an equivalent resistance of 4.615 ohms. (rounded). The 10 in series makes it 14.615...
5 0
3 years ago
Read 2 more answers
An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist
Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0
3 years ago
Why might it be necessary to ignore some of the data points just before and just after the collision?
krek1111 [17]
We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
7 0
3 years ago
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