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2 years ago

A person running has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?

Physics

1 answer:

allsm [11]2 years ago

7 0

Explanation:

Momentum is mass times velocity.

p = mv

720 kg m/s = m (5 m/s)

m = 144 kg

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Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

2 years ago

What is the equation: If F=10 N, a=5 m/s², m=?

Romashka-Z-Leto [24]

Answer:

2 kg

Explanation:

Remember:

F = m * a re-arrange to

F/a = m substitute in the given values

10 / 5 = 2 kg

8 0

2 years ago

Two masses are attracted by a gravitational force of 7 N.

damaskus [11]

Answer:

The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ? if the masses is quartered

Formula

$F = \frac{Km1m2}{r2}$

Process

Normal conditions F = Km₁m₂/r² = 7

When masses quartered F = K(m₁/4)(m₂/4)/r² = ?

F = K(m₁m₂/16)/r²

F = K(m₁m₂/16r² = 7/16 = 0.4375 N

3 0

2 years ago

What is the total distance that the object traveled?

Natasha_Volkova [10]

Answer:

for what?

Explanation:

d=S x T

d=vt+1/2at2

srry if wrong but

hope this helps

take care

8 0

3 years ago

2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .

muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4 mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28 x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

= .28 x 5/2 R x ( 620 - T )

= .28 x 5/2 x 8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

7 0

3 years ago