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telo118 [61]
3 years ago
11

When all the forces on an object are equal and opposite, & the net force is zero, that's..

Physics
1 answer:
Serggg [28]3 years ago
3 0

That would be a balanced force because the forces cancel each other out.

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Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ
bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c,  s= 400 m

r= ?

we have to solve for the radius

we know that circumference

c= 2\pi r

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular  displcament

∅= 400/63.63

∅= 6.28 radians

8 0
3 years ago
A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe
pentagon [3]

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0
3 years ago
Calculate the Latent Heat of Vaporization. (Please see picture attached)
Hunter-Best [27]

Answer:

20 J/g

Explanation:

In this question, we are required to determine the latent heat of vaporization

  • To answer the question, we need to ask ourselves the questions:

What is latent heat of vaporization?

  • It is the amount of heat required to change a substance from its liquid state to gaseous state without change in temperature.
  • It is the amount of heat absorbed by a substance as it boils.

How do we calculate the latent heat of vaporization?

  • Latent heat is calculated by dividing the amount of heat absorbed by the mass of the substance.

In this case;

  • Mass of the substance = 20 g
  • Heat absorbed as the substance boils is 400 J (1000 J - 600 J)

Thus,

Latent heat of vaporization = Quantity of Heat ÷ Mass

                                             = 400 Joules ÷ 20 g

                                             = 20 J/g

Thus, the latent heat of vaporization is 20 J/g

7 0
3 years ago
6. And what's up with ChiN?
n200080 [17]
I don’t know what’s up with chin man ??
4 0
2 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
3 years ago
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