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Mnenie [13.5K]
3 years ago
12

3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?

Physics
1 answer:
makkiz [27]3 years ago
7 0
I think it is B. 100 W
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What happens when a moving object bumps into a stationary object?.
Sonbull [250]
It depends on the mass of the moving object versus the mass of the stationary object. if the mass of the moving object is larger the stationary object will get sent into motion. if the mass of the stationary object is larger than the moving object, the stationary object will stay stationary and cause the moving object to do the same. if the two objects have the same mass, they will likely move together upon impact and then eventually come to rest.
5 0
2 years ago
What is the force needed to accelerate a wagon with a mass of 10 kg at a rate
Mashutka [201]

Answer:

B

Explanation:

Remember:

<u>F = m * a </u>

 =  10 * 2  = 20 N

6 0
2 years ago
Read 2 more answers
Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph .
Elis [28]
<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours

So he saves the time t_{s}:

t_{s}=t_{1}-t_{2}=2.07-1.85=0.22 \ hours

In minutes:

t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}

5 0
3 years ago
A 200 g air-track glider is attached to a spring. The glider is pushed in 9.8 cm against the spring, then released. A student wi
Likurg_2 [28]

Answer:

k =  5.05 N/m

Explanation:

In order to calculate the spring mass of the system, you use the following formula:

T=2\pi \sqrt{\frac{m}{k}}     (1)

T: period of oscillation of the system

m: mass of the air-track glider = 200g = 0.200 kg

k: spring constant = ?

You first calculate the period of oscillation:

T=\frac{1}{f}=\frac{1}{12/15.0s}=1.25s

Next, you solve the equation (1) for k, and then you replace the values of the other parmateres:

k=4\pi^2 \frac{m}{T^2}\\\\k=4\pi^2 \frac{0.200kg}{(1.25s)^2}=5.05\frac{N}{m}

The spring constant of the spring is 5.05 N/m

6 0
3 years ago
I WILL MARK YOU IF YOU HELP
sertanlavr [38]
F=ma
11.6=3.8*a
a=11.6/3.8
a=3.05m/s
6 0
2 years ago
Read 2 more answers
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