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EastWind [94]
3 years ago
7

What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)

Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=<em>k</em>ma (where k is the proportionality constant)

<em>k</em> is 1 so you can ignore it.

So, our equation becomes F=ma

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It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0
VikaD [51]
The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have 

rate = \frac{-(C_{new} - C_{old})}{t}

where C are the concentration values and t is the time taken for it to decompose.

14 = \frac{-(0.02 - 0.06)}{t}
t = 0.003 s

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0
3 years ago
A girl pulls on a 10-kg wagon with a constant horizontal force of 30 N. If there are no other horizontal forces, what is the wag
strojnjashka [21]
Force equals mass*distance
F = ma

Given m = 10 kg, F = 30 N

30 = 10a
30/10 = a
3 = a

The wagon's acceleration is 3 m/s^2
7 0
3 years ago
A bullet starting from rest accelerates uniformly at a rate of 1,250 meters per square second. What is the bullet's speed after
Tatiana [17]

Answer:

125,000

Explanation:

8 0
3 years ago
80
Dahasolnce [82]

Answer:

1 million hahahahahahahahhahah

7 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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