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pshichka [43]
2 years ago
5

If a force of 10 N acts on an object and an additional force of 6 N acts on the object in the same direction. What will be the n

et force be on the object?
A- 60 N
B- 16 N
C- 6 N
D- 4 N

Use Newton's 2nd law (f=ma) to solve the following problem.

Grady pushed his tennis ball with a mass of 30 kg causing an acceleration of 1 m/s/s. What force is Grady applying?
A- 30 N
B- 30
C- There is no force, the tennis ball did not move
D- 1 N

Physics
2 answers:
NISA [10]2 years ago
8 0
First is B second is A third is D
Schach [20]2 years ago
4 0

Answer: The 1st one is B. I'm pretty sure the 2nd one is A. and the 3rd one is D.

Explanation: I hope this helped, although I'm not sure for the second one

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A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

5 0
3 years ago
A person is riding a motorized tricycle. They weigh 180kg and are moving at 3 m/s over a distance of 300 m. How much work is don
agasfer [191]

If I am to understand this question correctly this is what asks you:

If a person is riding a motorized tricycle how much work do they do?

You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any <em><u>motorized vehicle</u></em> other than the time it takes to get from one place to another.

So to answer this question I would say:

Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.

If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!

3 0
3 years ago
After a magnetic field is applied, how many energy levels would you expect for then=3,ℓ=2 state, including electron spin?
zloy xaker [14]

Answer:

The energy level is 5.

Explanation:

Given that,

n = 3

l = 2

We know that,

l shows the number of sub-shells and define the number of angular nodes.

n shows the number of electron shell.

m_{l} is a quantum number. It is define the number of energy level in a sub-shells .

m_{s} is define the spin of the electron.

So, The quantum number is

m_{l}=-2,-1,0,1,2

m_{s} is +\dfrac{1}{2} and -\dfrac{1}{2} for every energy level.

The energy level is 5.

Hence, The energy level is 5.

7 0
3 years ago
What is mean by combination reaction ?​​
mrs_skeptik [129]

\underline{\purple{\large \sf Combination \: reaction :-}}

Those reaction in which two or more substances combine to form a one new substance are called Combination reaction

In this reaction, We can add :

  • Two or more elements can combine to form a compound.
  • Two or more compounds can combine to from a one new compound.
  • An element and a compound can combine to form a new compound.

\underline{\green{\large \sf For\: example :}}

\sf 2H_{2}  + O_{2} \:  \underrightarrow{Combination} \: 2H_{2}o

In this, Hydrogen is an element and Oxygen is another element. Both are combined to form compound 'Hydrogen oxide'. Hydrogen oxide is commonly known as water.

3 0
3 years ago
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C
Alex_Xolod [135]

Answer:

26.466cm³/min

Explanation:

Given:

Volume 'V'= 320cm³

P= 95kPa

dP/dt = -11 kPa/minute

pressure P and volume V are related by the equation

PV^{1.4}=C

we need to find dV/dt, so we will differentiate the above equation

V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt}  = \frac{d[C]}{dt}

\frac{dP}{dt} V^{1.4} + P(1.4)V^{0.4} \frac{dV}{dt} =0

lets solve for dV/dt, we will have

\frac{dV}{dt} =\frac{-\frac{dP}{dt} V^{1.4} }{P(1.4)V^{0.4} } \\\frac{dV}{dt} =- \frac{-\frac{dP}{dt} V}{P(1.4)}

\frac{dV}{dt} = -\frac{(-11 ) 320}{95(1.4)}  (plugged in all the values at the instant)

\frac{dV}{dt} = 26.466

Therefore, the volume increasing at the rate of 26.466cm³/min at this instant

6 0
3 years ago
Read 2 more answers
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