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Jet001 [13]
2 years ago
12

Two objects of mass 3 kg and 2 kg are thrown with velocities 2 ms-1 and 3 ms-1 , respectively. Calculate the momentum of both. W

hich has a greater momentum?​
Physics
1 answer:
Ludmilka [50]2 years ago
4 0

Answer:

They have the SAME momentum ....see below

Explanation:

Momentum = m v

3       3kg * 2 m/s =   6  kg-m/s

2        2kg * 3 m/s =   6 kg - m/s        

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Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.
galina1969 [7]
1 in=2.54 cm=(2.54 cm)(1 m/100 cm)=0.0254 m
Therefore:
1 in=0.0254 m
1 in³=(0.0254 m)³=1.6387064 x 10⁻⁵ m³

Therefore:

8.06 in³=(8.06 in³)(1.6387064 x 10⁻⁵ m³ / 1 in³)≈1.321 x 10⁻⁴ m³.

Answer: 8.06 in³=1.321 x 10⁻⁴ m³
8 0
3 years ago
List 3 cases in which potential energy becomes kinetic energy and three cases in which kinetic energy becomes potential energy.
Mrrafil [7]

Answer:

when you open a can of pop

when you jump on your bed

Explanation:

3 0
3 years ago
three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0
3 years ago
How much ice (at 0°C) must be added to 1.90 kg of water at 79 °C so as to end up with all liquid at 8 °C? (ci = 2000 J/(kg.°
muminat

m = mass of the ice added = ?

M = mass of water = 1.90 kg

c_{w} = specific heat of the water = 4186 J/(kg ⁰C)

c_{i}  = specific heat of the ice = 2000 J/(kg ⁰C)

L_{f} = latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg

T_{ii}  = initial temperature of ice = 0 ⁰C

T_{wi} = initial temperature of water = 79 ⁰C

T = final equilibrium temperature = 8 ⁰C

using conservation of heat

Heat gained by ice = Heat lost by water

m c_{w}  (T - T_{ii} ) + m L_{f}  = M c_{w}  (T_{wi} - T)

inserting the values

m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)

m = 1.53 kg

4 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
PtichkaEL [24]

This is True

Kinetic energy is the energy of motion. The bicyclist is in motion as he pedals up the tall hill. Therefore, the bicyclist contains kinetic energy.

4 0
4 years ago
Read 2 more answers
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