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2 years ago

7) HELP ASAP PLZ PLZ HELP ME I don't know if I am right so if I am not right can you plz help me

Physics

2 answers:

artcher [175]2 years ago

6 0

It’s either “B” or “A”

CaHeK987 [17]2 years ago

5 0

I believe that the answer is B

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10. What is the acceleration of a 1000 kg car subject to a 500 N net force?

Simora [160]

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

3 0

1 year ago

A kettle full of water is brought to a boil in a room with temperature 20Â°c. after 15 minutes the temperature of the water has

kogti [31]

Answer: The temperature after another 5 minutes is 68.5Â°c

Explanation: Please see the attachments below

4 0

2 years ago

Gerry is looking at salt under a powerful microscope and notice a crystalline structure.what can be known about the salt sample

Natali [406]

Well, if the salt that Gerry's looking at under a powerful microscope has a crystalline structure, then that's saying that salt is technically a solid.

(I hope that this is an answer you were looking for)

6 0

2 years ago

Find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the

Natali [406]

Answer:

The potential energy of the hiker is $1.13\times 10^6\ J$ .

Explanation:

Given that,

Mass of the hiker, m = 61 kg

Height above sea level, h = 1900 m

We need to find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington. The potential energy is given by :

$E=mgh$

g is the acceleration due to gravity

$E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J$

So, the potential energy of the hiker is $1.13\times 10^6\ J$ . Hence, this is the required solution.

8 0

2 years ago

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

4 0

2 years ago