Answer:
m=0.5kg
h = 180 cm =1.8 mh=180cm=1.8m
Initial potential energy of the object is:
E_p=m*g*hE
p
=m∗g∗h
Kinetic energy at the surface:
E_k=\frac{mv^2}{2}E
k
=
2
mv
2
According to the law of conservation of energy (assuming no air resistance):
E_p = E_kE
p
=E
k
mgh=\frac{mv^2}{2}mgh=
2
mv
2
Solving for v:
v=\sqrt{2gh}v=
2gh
p=mvp=mv
So,
p= m*v = m\sqrt{2gh}p=m∗v=m
2gh
Calculating:
p= 0.5\sqrt{2*9.8*1.8}\approx 2.97 \frac{kg*m}{s}p=0.5
2∗9.8∗1.8
≈2.97
s
kg∗m
Answer:
p \approx 2.97 \frac{kg*m}{s}p≈2.97
s
kg∗m
Answer:
a projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
Explanation:
Answer:
<h3>The answer is 0.92 m/s²</h3>
Explanation:
To find the acceleration of an object given it's initial and final velocity and time taken we use the formula
![a = \frac{v - u}{t} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bv%20-%20u%7D%7Bt%7D%20%20%5C%5C%20)
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question
v = 39 m/s
u = 27 m/s
t = 13 s
We have
![a = \frac{39 - 27}{13} = \frac{12}{13} \\ = 0.923076... \: \: \:](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B39%20-%2027%7D%7B13%7D%20%20%3D%20%20%5Cfrac%7B12%7D%7B13%7D%20%20%5C%5C%20%20%3D%200.923076...%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20)
We have the final answer as
<h3>0.92 m/s²</h3>
Hope this helps you
Answer:
The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
Explanation:
In an electric field, a positive charge has tendency to move from high to low potential and hence experience the electric force in the direction of electric field since electric field lines are directed from high to low potential.
In an electric field, a negative charge has tendency to move from low to high potential and hence experience the electric force in the direction opposite to electric field since electric field lines are directed from high to low potential.This phenomenon happened because The electric field from a positive charge will points away from the charge while the electric field from a negative charge will points toward the charge.