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Charra [1.4K]
3 years ago
12

Problems - Show all work.

Physics
1 answer:
Fittoniya [83]3 years ago
4 0

Answer:

21s

Explanation:

Given parameters;

Radius  = 10m

Speed or velocity  = 3m/s

Unknown:

Period  = ?

Solution:

To solve this problem, use the expression:

      v  = \frac{2\pi r}{T}  

r is the radius

T is the unknown

           Input the parameters and solve for T;

    3  = \frac{2 x \pi  x 10}{T}  

    62.84 = 3T

         T  = 21s

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If the car has a mass of 0.4 kg, the ratio of height to width of the ramp is 19/85, the initial displacement is 1.9 m, and the c
Rasek [7]

Answer:

final displacement lf  = 0.39 m

Explanation:

from change in momentum equation:

\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]

given: m = 0.4kg, y/x = 19/85, li = 1.9 m,

\delta p = 1.27 kg*m/s.

putting all value to get the final displacement value

1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]

final displacement lf  = 0.39 m

5 0
3 years ago
If the temperature of the metal oxide and water solution is increased,this would most likely..
kiruha [24]
Chemical Reaction between metal oxide and water solution
7 0
3 years ago
An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)
Alecsey [184]
"60 kg" is not a weight.  It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is   

                                 (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

                   Weight = (60 kg) x (9.8 m/s²)

                                =      588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to  5R  from the center, the gravity out there is

                    (1R/5R)²  =  (1/5)²  =  1/25  =  0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

                 (0.04) x (588 Newtons)  =  23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
6 0
3 years ago
Help me &lt;3 please <br> Thank you :)
NARA [144]

Answer:

11,890

Explanation:

First we need to know what is considered a significant figure.

A significant figure is a value that is not a zero at the start OR end of a value.

Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.

The 0 in the value of 3056 is considered a significant figure.

So from the table, we can deduce:

0.275 has 3 significant figures

750 has 2 significant figures

10.4 \times  {10}^{5}  = 1040000

has 3 significant figures.

11,890 has 4 significant figures.

320,050 has 5 significant figures.

So from the above, we can already see the answer.

4 0
2 years ago
Answer the question fast please​
AnnZ [28]

Answer:

1

Explanation:

d = m \div v  \\ m = 0.8 \: v = 80\% \: also \: 0.8 \\ d = 0.8 \div 0.8 \\ d = 1

6 0
3 years ago
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