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KATRIN_1 [288]
3 years ago
6

if a 60 kg person was standing on a platform at the surface of saturn and they jumped, they would have to push with a force grea

ter than..?

Physics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

A 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N

Explanation:

The gravitational attraction between an object on the surface of a planet and the planet is given by the weight of the object

Therefore the force needed to be applied for an object to lift off the surface of a planet = The weight of the object

The weight of the object on the surface of a planet = m × g

Where;

m = The mass of the object

g = The strength of gravity on the planet's surface in N/kg

The given parameters are;

The mass of the person standing on a platform at the surface of Saturn, m = 60 kg

The strength of gravity on the surface of Saturn = 9 N/kg

Therefore, we have;

The weight of the person = The force greater than which the person would have to push on the surface of Saturn so as to Jump = The weight of the person on the surface of Saturn = 60 kg × 9 N/kg = 540 N

Therefore, for a 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N.

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a bus travels 4 km due north and 3 km due west going from bus station a to bus station b. the magnitude of the bus displacement
ladessa [460]

Answer:

5km

Explanation:

Magnitude of displacement is found by getting the resultant. Resultant is same as the bypotenuse hence

R=\sqrt {{x^{2}+y^{2}} where x is the displacement in west direction and y is displacement in North direction. Substituting x with 3km and y with 4 km then

R=\sqrt {{3^{2}+4^{2}}=5km

4 0
3 years ago
How can the direction of a tensional force be changed without diminishing the force?
RSB [31]

Given what we know, we can confirm that the tensional force of a system can in theory be changed without diminishing its force through the use of an ideal pulley.

<h3>What is an ideal pulley?</h3>
  • A pulley is a small wheel through which a string or chain is run.
  • These are used in order to change the direction of a force.
  • An ideal pulley would be one in which there is no friction and the pulley itself would have no mass.
  • Therefore, the force would be able to change directions without giving part of its force to the pulley system.

Therefore, we can confirm that the only known way to change the direction of a force without diminishing its value would be through the use of a frictionless and massless pulley system otherwise known as an ideal pulley.

To learn more about Friction visit:

brainly.com/question/13357196?referrer=searchResults

4 0
2 years ago
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
4 years ago
ANYONE who is good with the consequences of Population growth PLEASE help me!!
alekssr [168]

in china, there is a family limit for only having 1 child

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4 0
3 years ago
Read 2 more answers
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
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