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Anarel [89]
3 years ago
12

What is the answer? please

Physics
2 answers:
Nitella [24]3 years ago
7 0
I think it`s D but iam not so sure 
anastassius [24]3 years ago
6 0
The third one down is the answer.
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What is the net force on an object that has a mass of 2.5 kg and accelerates to the right at 2 m/s?
Nadya [2.5K]

Answer:

5 N

Explanation:

The formula that you've to use to find the force is :

Force = Mass × Acceleration

They have already given that,

m = 2.5 kg

a = 2 ms⁻²

Let us find now.

F = m a

F = 2.5 kg ×  2 ms⁻²

F =  5 N

Hope this helps you :-)

Let me know if you have any other questions :-)

5 0
2 years ago
A hot-air balloon is floating above a straight road. To estimate their height above the ground, the balloonists simultaneously m
ki77a [65]
We know that
tan(∅) = y/x; where y will be the height of the balloon and x will be the distance from the milestone marker.
The marker that is further away will produce a smaller angle of depression.
tan(18) = height / x
x = height / tan(18)

The second marker is x + 1 miles away:
tan(16) = height / x + 1; substituting x:
tan(16) = height / (height/tan(18) + 1)
tan(16) x [(height / tan(18) + 1] = height
0.88h + 0.29 = h
h = 2.42 miles
8 0
3 years ago
For the population, scores on the test are normally distributed with μ = 70 and σ = 15. The sample of n = 25 students had a mean
AysviL [449]

Answer:

z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67      

p_v =2*P(t_{24}>1.67)=0.108  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.    

Explanation:

1) Data given and notation      

\bar X=75 represent the sample mean

\sigma=15 represent the standard deviation for the population

n=25 sample size      

\mu_o =70 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the mean is different from 70, the system of hypothesis would be:      

Null hypothesis:\mu = 70      

Alternative hypothesis:\mu \neq 70      

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:      

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)      

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic      

We can replace in formula (1) the info given like this:      

z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

df=n-1=25-1=24  

Since is a two tailed test the p value would be:      

p_v =2*P(t_{24}>1.67)=0.108  

Conclusion      

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.      

3 0
4 years ago
A rock dropped down a well takes 1.8 s to hit the water. How far below the top of the well is the surface of the water
Brut [27]

Answer: 15.87 m

From the equation of motion:

s=ut+\frac{1}{2}at^2

where, s is the distance traveled, u is the initial velocity, a is the acceleration and t is the time.

The rock free falls under gravity. Initial velocity, u=0 m/s, a=g=9.8m/s^2

It took t=1.8 s for rock to hit the water.

Substitute the values in the given equation:

\Rightarrow s=0+\frac{1}{2}9.8m/s^2\times(1.8s)^2=15.87 m

Hence, the water is 15.87 m below the top level of the well.

3 0
4 years ago
Why would it be incorrect to say that the bulb supplies 60 J of light energy energy each second
LekaFEV [45]

Answer:

Explanation:

A 60W bulb , consumes electrical energy of 60 J per second but it does not give out all the energy in the form of light . Most of the energy consumed by it is wasted in the form of heat . Only a fraction of energy is converted into light energy .

Hence, it is wrong to say that bulb supplies 60 J of light energy each second .

4 0
3 years ago
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