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olchik [2.2K]
3 years ago
14

Insta: EthanbCaldwell

Physics
2 answers:
Oliga [24]3 years ago
7 0

Answer:

<h2><u><em>ok</em></u></h2>

Explanation:

olasank [31]3 years ago
5 0

Answer:

XD ok m8

Explanation:

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What is the metric unit of weight? Gram/pound/Newton/milliliter
coldgirl [10]
Weight is force.
The metric unit of force is the Newton.
1 Newton is the force needed to accelerate 1 kilogram at the rate of 1 m/s² .
7 0
3 years ago
On the surface of planet y, which has a mass of 4.83x1024 kg, a 30.0 kg object weighs 50.0 n. what is the radius of the planet?
Aleksandr [31]
We can solve the problem in two steps:

1) From the weight W=50.0 N of the object, we can find the value of the gravitational acceleration g of the planet. In fact, the weight is equal to
W=mg
where m=30 kg is the mass of the object. From this, we find g:
g= \frac{W}{m}= \frac{50.0 N}{30 kg}=1.67 m/s^2

2) The gravitational acceleration of a planet with mass M and radius r is given by
g= \frac{GM}{r^2}
where G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational  constant. In our problem, the mass of the planet is 
M=4.83 \cdot 10^{24} kg, and we found g in step 1), g=1.67 m/s^2, so we have everything to solve and find the value of the radius r:
r= \sqrt{ \frac{GM}{g} }= \sqrt{ \frac{(6.67\cdot 10^{-11})(4.83 \cdot 10^{24})}{1.67} }=1.39\cdot 10^7 m
5 0
3 years ago
Good experimental design includes a hypothesis that is
Ratling [72]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Good experimental design includes a hypothesis that is <span>testable and observable.</span>
3 0
4 years ago
Read 2 more answers
A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0
4 years ago
Read 2 more answers
A ball is dropped from a 100-m tall building. Neglecting air resistance, what will the speed of the ball be when it reaches the
VARVARA [1.3K]
V=u+at=0+(9.8)(4.5)=44.1
5 0
3 years ago
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