All categories

3 years ago

Can somebody help me on finding the constants if this experiment?

Physics

1 answer:

ki77a [65]3 years ago

8 0

The constants could be the day because the experiment is all on the same day or the person throwing the pumpkin because it will always be the same perosn

You might be interested in

Find B when θ=35, E=10V, t=5, N=250, A=1.20m^2

givi [52]

Answer:

its most definitely c. trust me

Explanation:

3 0

2 years ago

Scientific ideas about the solar system have changed over time. Which of them

Usimov [2.4K]

Answer:

Explanation:

If i'm not wrong and late it might be F

7 0

2 years ago

You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha

UNO [17]

V = IR

By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.

So i believe the answer would be : The string will remain lit

hope this helps

6 0

3 years ago

A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg

ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

$Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3$

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

$p=m/V\\\\ p=m/(L*A)\\p*A=m/L$

Now the equation for speed of traverse wave is calculated through:

$\sqrt \frac{F*L}{m}\\$

= $\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}$

Substituting the values

$\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3} \\$

=22.49 m/s

4 0

3 years ago

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0

3 years ago