Ask question

Ask question

All categories

1 year ago

5

a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the

moon, where the acceleration due to gravity is determine to three significant figures its weight in si units and its mass in si units

1 answer:

Katena32 [7]1 year ago

7 0

Answer:

$W_{earth} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 5.91 * 10^6 N$ : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

$m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg$

$g_{moon}= 1.62 \frac{m}{s^2}$

$g_{earth}= 9.8 \frac{m}{s^2}$

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

$W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N$ : Rocket weight on moon

You might be interested in

One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi

Aneli [31]

Answer:

Explanation:

Given

initial speed(u)=5 m/s

Final speed(v)=4 m/s

Distance traveled=3 m

using equation of motion

$v^2-u^2=2as$

$4^2-5^2=2(a)(3)$

$a=\frac{-3}{2}=-1.5 m/s^2$

after this its final velocity will be zero

$v^2-u^2=2as$

$0^2-4^2=2\times (-1.5)\times s$

s=5.33 m

Total distance=3+5.33=8.33 m

Thus he will not be able to win the game

4 0

1 year ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

Flauer [41]

Answer:

The new resistance becomes half of the initial resistance.

Explanation:

The resistance of a wire is given by :

$R=\dfrac{\rho L}{A}$

$\rho$ = resistivity of material

L and A are linear dimension

If the electrical wire is replaced with one having every linear dimension doubled i.e. l' = 2l and r' = 2r

New resistance of wire is given by :

$R'=\dfrac{\rho L'}{A'}$

$R'=\dfrac{\rho (2L)}{\pi (2r)^2}$

$R'=\dfrac{1}{2}\dfrac{\rho L}{A}$

$R'=\dfrac{1}{2}R$

The new resistance becomes half of the initial resistance. Hence, this is the required solution.

4 0

2 years ago

A radioactive substance decays exponentially. A scientist begins with 200 milligrams of a radioactive substance. After 17 hours,

lianna [129]

75.17 mg of the radioactive substance will remain after 24 hours.

Answer:

Explanation:

Any radioactive substance will obey the exponential decay behavior. So according to this behavior, any radioactive substance will be decaying in terms of exponential form of disintegration constant and Time.

Disintegration constant is the rate of decay of radioactive elements. It can be measured using the half life time of the radioactive element .While half life time is the time taken by any radioactive element to decay half of its concentration. Like in this case, at first the scientist took 200 mg then after 17 hours, it got reduced to 100 g. So the half life time of this element is 17 hours.

Then Disintegration constant = 0.6932/Half Life time

Disintegration constant = 0.6932/17=0.041

Then as per the law of disintegration constant:

$N = N_{0}e^{-xt}$

Here N is the amount of radioactive element remaining at time t and $N_{0}$ is the initial amount of sample, x is the disintegration constant.

So here, $N_{0}$ = 200 mg, x = 0.041 and t = 24 hrs.

N = 200 × $e^{-24*0.041}$ =75.17 mg.

So 75.17 mg of the radioactive substance will remain after 24 hours.

3 0

1 year ago

What is the magnitude of the relative angle φ

melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0

1 year ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0

1 year ago