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IrinaK [193]
4 years ago
10

James (mass 85.0 kg) and Ramon (mass 69.0 kg) are 20.0 m apart on a frozen pond. Midway between them is a mug of their favorite

beverage. They pull on the ends of a light rope stretched between them. Ramon pulls on the rope to give himself a speed of 1.10 m/s.What is James's speed?
Physics
1 answer:
-BARSIC- [3]4 years ago
7 0

Answer:

v = –0.89m/s.

Explanation:

This problem require the knowledge of conservation of momentum

Given the mass of James and Ramon to be 85kg and 69kg respectively.

Total initial momentum = 0 since James and ramon were both not moving initially.

Ramon has a final momentum = 69.0 * 1.10

James has a final momentum = 85 * vf

85×v + 69×1.10 = 0

85v + 75.9 =

85v = –75.9

v = –75.9/85 = –0.89m/s

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Calculate the pressure on a man’s foot when a woman of mass 65 kg steps on his foot with her heel which has an area of 0.12m2 wi
tino4ka555 [31]

Answer:

Pressure = force/area

= 650N/0.12m^2

= 5416.7 to 1 d.p

I got 650N by taking the gravitational field on earth to be 10N/kg.

Weight = mass x gravitational field

= 65 x 10N/kg

= 650

Answer is 5416.7 pascals (Pa)

Or

5308.3 Pa

if gravitational field was taken to be 9.8N/kg

Hope this helps!

4 0
3 years ago
an airplane maintains a speed of 567 km/h relative to the air it is flying through as it makes a tripto a city 780 km away to th
Vera_Pavlovna [14]
A ) v = 567 - 33 = 534 km/h
     v = d / t    ⇒   t = d / v
     t = 780 km / 534 km/h = 1.46 h 
b ) v = 567 + 33 = 600 km/h
     t = 780 km / 600 km/h = 1.3 h
c ) v = (567² +33²) ^(1/2) = √320,400 = 566 km/h
     t = 780 km/ 566 km/h = 1.378 h
4 0
3 years ago
Two heavy objects of masses 1 kg and 2 kg are moving towards each other under
SOVA2 [1]

The acceleration of the body in terms of the gravitational constant G is G.

According to Newton's law of universal gravitation;

F = Gm1m2/r^2

G = gravitational constant

m1 = mass of the first body

m2 = mass of the second body

r = distance between the two bodies

Substituting values to find the force on the two bodies;

F = G × 1  × 2/1^2

F = 2G

For the 2 Kg mass

F = ma

m = mass

a = acceleration

F = gravitational force

Hence,

2G = 2a

a = 2G/2

a = G

Learn more: brainly.com/question/13860566

5 0
3 years ago
A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

⠀

⠀

\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

⠀

⠀

\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

⠀

⠀

<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

⠀

\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

⠀

⠀

\large\textsf{\underline{Now substituting the required values}}

⠀

⠀

\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

⠀

\textsf{ \large {\underline{Now substituting the required values}}  }

⠀

⠀

\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

⠀

⠀

⠀

<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

8 0
2 years ago
An object moving with constant acceleration near the surface of the earth
ziro4ka [17]
The answer is C because if it’s moving with a constant speed it’s gonna be balanced and there will be no force
6 0
3 years ago
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