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4vir4ik [10]
2 years ago
10

Amber moves from her desk to the door in 17 seconds. Her speed changes from rest to 5 m/s. What is Amber's average acceleration?

Physics
1 answer:
VMariaS [17]2 years ago
4 0
Vf - vi/ time
5 - 0/17
0.29 m/s square
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An athlete runs a 100 m race in 10 seconds against a friction force of 100N. How do I work out his power output?
Rashid [163]
They seem to cancel each other out which is odd
6 0
3 years ago
A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th
Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m

6 0
3 years ago
Tara prepared a report to show how the amplitude of waves affects the energy of waves. Is her graphical representation correct?
Gelneren [198K]

Answer:

It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down

Explanation: I got the question right

6 0
3 years ago
Read 2 more answers
Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0. 250 kilograms, what should be
antoniya [11.8K]
Sum of all forces = mass * acceleration

Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)

Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
Ft= 2.45N

7 0
2 years ago
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
3 years ago
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