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2 years ago

Please help i will mark brainliest

Physics

2 answers:

valkas [14]2 years ago

6 0

Im pretty sure the answer is c; velocity is usually measure by a vector :)

yarga [219]2 years ago

4 0

Answer:

c. speed

Explanation:

Speedometer is a device used to measure the speed of a vehicle. I am pretty sure this is the correct option.

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The input work done on a machine is 9.63 × 103 joules, and the output work is 3.0 × 103 joules. What is the percentage efficienc

notsponge [240]

Input work = 9.63×10³ J.
Output work = 3.0×10³ J

By definition,
Efficiency = (Output work)/(Input work)
= (3.0×10³)/(9.63×10³)
= 0.31 = 31%

Answer: 31%

5 0

3 years ago

Read 2 more answers

In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the

sp2606 [1]

<span>I'll tell you how to do it but you must crunch the numbers.

Use Kepler's 3rd Law

T^2 = k R^3

where k = 4(pi)^2/ GM
G =gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2
M = mass of this new planet
pi = 3.14159265
T =3.09 days = 266976 seconds
R = (579,000,000km)/9 = 64333333.3 km

a)
Solve Kepler's 3rd Law for M. Your answer will be in kg

b)
mass of the sun = 1.98892 × 10^30 kilograms
Form the ratio
M(planet)/M(sun) </span>

6 0

2 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

The orbital speed is $v= 2.6*10^{3} m/s$

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at

stira [4]

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

Angular speed ,ω = 150 rpm

The speed in rad/s

$\omega =\dfrac{2\pi N}{60}$

$\omega =\dfrac{2\pi \times 150}{60}$

ω = 15.7 rad/s

The distance of center of mass from 150 g

$r=\dfrac{150\times 0+240\times 33}{150+240}\ cm$

r= 20.30 cm

The speed of the mass 150 g

v= ω r

v= 20.30 x 15.7 cm/s

v= 318.71 cm/s

v= 3.18 m/s

6 0

2 years ago

Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E

Leviafan [203]

We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
$\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}$
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2$
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
$mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\$
Because both carts have the same mass we can cancel those out:
$3v=v_A+2v_B$
Now we have our system of equation that we have to solve:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B$
Part A
We need to solve our system for $v_a$ . We will solve second equation for $v_b$ and then plug that in the first equation.
$3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}$
Now we have to plug this in the first equation:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
We will multiply the first equation with 2 and divide by m:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
Now we plug in the second equation into first one:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0$
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
$a=1\\
b=-6v\\
c=v^2-\frac{E}{m}$
Solutions are:
$v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}$
Part B
We do the same thing here, but we must express $v_a$ from momentum equation:
$3v=v_A+2v_B\\
v_A=3v-2v_B$
Now we plug this into our energy conservation equation:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\
3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\
3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\
3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\
3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\
3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0
$
Again we end up with quadratic equation. Coefficients are:
$a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}$
Solutions are:
$v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}$

8 0

2 years ago