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earnstyle [38]
2 years ago
15

An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°

Physics
1 answer:
Y_Kistochka [10]2 years ago
4 0
Becaused it the thing was cod that why
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A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c
VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0
2 years ago
If you can't throw a football what do u need to work on more
NeX [460]
Work out your arms.                                                          
3 0
3 years ago
Read 2 more answers
A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?
nata0808 [166]

Answer:

v=u+at

v=21+1.4(11)

v=21+15.4

v=36.4

3 0
2 years ago
If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a result?
Fittoniya [83]

Answer:

The Kinetic Energy decreases. The Total Energy stays the same

Explanation:

The TE stays the same, so if PE increases then KE will decrease.

7 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
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