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Pavel [41]
3 years ago
15

What net external force is required to give a 25 kg suitcase an acceleration of 2.2 m/s2 to the right?

Physics
1 answer:
lapo4ka [179]3 years ago
3 0

Answer:

there are two way to get mate  and i gave them sepaert explation

Explanation:

55

N

Explanation:

Using Newton's second law of motion:

F

=

m

a

Force=mass

×

acceleration

F

=

25

×

2.2

F

=

55

N

So 55 Newtons are needed.

Answer link

Nam D.

Apr 6, 2018

55

N

Explanation:

We use Newton's second law of motion here, which states that,

F

=

m

a

m

is the mass of the object in kilograms

a

is the acceleration of the object in meters per second

F

=

25

kg

⋅

2.2

m/s

2

=

55

N

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A block of ice (m = 9 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of t
jonny [76]

Answer:

a) An expression for the amount of energy, E_m, needed to melt the ice into water.

(E_m) = (m × Lf)

b) An expression for the total amount of energy, E_tot, to melt the ice and then bring the water to T2

(Total heat) = (m × Lf) + mc (T2 - T1)

c) 3,646,458 J = 3646.46 kJ

Explanation:

a) When a pure body changes its phase at meltimgbor boiling point, it does so at a constant temperature. When a pure body melts, the amount of heat responsible for this change is just given by a product od the mass of the body and the body's heat of fusion.

(E_m) = (m × Lf)

b) The Heat required to raise the temperature of a body from one temperature to another is given by the product of the mass of the body, its specific heat capacity and the temperature difference between the final point and the starting point.

(E_2) = mcΔT = mc (T2 - T1)

Total heat required to melt the ice at T1 = 0 and raise the temperature of the resulting water to T2 is then a sum of (E_m) + (E_2)

(Total heat) = (m × Lf) + mc (T2 - T1)

c) What is the energy in Joules?

(Total heat) = (m × Lf) + mc (T2 - T1)

m = mass of ice = resulting mass of water = 9 kg

Lf = latent heat of fusion = 334000 J/kg

c = Specific heat capacity of water = 4186 J/kg.K

T2 = final temperature of the water = 17°C

T1 = Initial temperature of the water = 0°C

Note that the units of temperature difference is the same for K and °C

(Total heat) = (m × Lf) + mc (T2 - T1)

Q = (9 × 334000) + [9 × 4186 × (17 - 0)]

Q = 3,006,000 + 640,458 = 3,646,458 J = 3646.46 kJ

Hope this Helps!!!

7 0
3 years ago
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