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3 years ago

13

Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would

weigh 34.65 Newtons on Jupiter, what is the strength of gravity there.

1 answer:

Andre45 [30]3 years ago

7 0

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

$W = mg$

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

$W_{x} = mg_{x}$

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

$34.65 N = (1.5 kg)g_{x}$

$g_{x} = \frac{34.65 N}{1.5 kg}$

<u>gₓ = 23.1 m/s²</u>

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if a bowling ball and a golf ball or move at the same velocity, which one would have more momentum? Why?

Vlada [557]

I would Say the Bowling Ball, Would have more momentum, Because of Newtons 3rd Law.

I hope I helped

- Dante

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3 years ago

Read 2 more answers

A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

7 0

3 years ago

A world class sprinter is travelling with speed 12.0 m/s at the end of a 100 meter race. Suppose he decelerates at the rate of 2

Solnce55 [7]

Answer:

after 6 second it will stop

he travel 36 m to stop

Explanation:

given data

speed = 12 m/s

distance = 100 m

decelerates rate = 2.00 m/s²

so acceleration a = - 2.00 m/s²

to find out

how long does it take to stop and how far does he travel

solution

we will apply here first equation of motion that is

v = u + at ......1

here u is speed 12 and v is 0 because we stop finally

put here all value in equation 1

0 = 12 + (-2) t

t = 6 s

so after 6 second it will stop

and

for distance we apply equation of motion

v²-u² = 2×a×s ..........2

here v is 0 u is 12 and a is -2 and find distance s

put all value in equation 2

0-12² = 2×(-2)×s

s = 36 m

so he travel 36 m to stop

3 0

3 years ago

16x^2y^2-25a^2b^2<br>factorize the expression

SIZIF [17.4K]

Answer:

(4xy+5ab)(4xy-5ab)

Explanation:

16 $x^{2}$ $y^{2}$ -25 $a^{2}$ $b^{2}$

4^2 is 16 and 5^2 is 25,

Also, (x-a)(x+a) = x^2-a^2

So, this factorized is:

(4xy+5ab)(4xy-5ab)

Hope this helps!

8 0

3 years ago

3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of

Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

$n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555$

Again using Snell's law for blue light as :

$n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283$

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0

3 years ago