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kap26 [50]
3 years ago
10

A projectile is launched straight upward at 50 m/s. Neglecting air resistance its speed upon returning to its starting point is:

________.
Physics
1 answer:
lbvjy [14]3 years ago
4 0

Answer:

The speed of the projectile upon returning to its starting point is 50 m/s.

Explanation:

Given;

initial velocity of the projectile, u = 50 m/s

The velocity of the projectile is maximum before hitting the ground. As the object moves upward, the velocity reduces as it approaches the maximum height, at the maximum height the velocity becomes zero. Also, as the object moves downward, the velocity starts to increase and becomes maximum before hitting the ground.

Therefore, the speed of the projectile upon returning to its starting point is 50 m/s.

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The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor
uysha [10]

Answer:

c.a_m

Explanation:

We are given that

Acceleration due to gravity on the moon=a_m

Acceleration due to gravity on the earth=a_e

g_m=\frac{1}{6}g_e

Net force due to am on an object on moon=F_{net}=ma_m

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=F=ma_e

F=F_{net}(given)

ma_m=ma_e

a_e=a_m

Hence, option c is true.

3 0
3 years ago
A friend claims that there are just two forms of energy: kinetic energy and gravitational potential energy. What is wrong with y
Oksanka [162]

Answer:

D

Explanation:

There is a variety of many types of energies. Some of which are mechanical, thermal, electromagnetic, nuclear, chemical, sonic, and more.

7 0
3 years ago
What is the net force on this object?
damaskus [11]

Upward and downward forces cancel out. Net force is 8 newtons to the right

5 0
3 years ago
Read 2 more answers
A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a
MAXImum [283]

Answer:21.18 m

Explanation:

Given

initial speed u=10 m/s

height of building h=22 m

time taken to complete 22 m

h=ut+\frac{1}{2}at^2

initial vertical velocity =0

22=\frac{1}{2}gt^2

t=\sqrt{\frac{22\times 2}{g}}

t=2.11 s

Horizontal Distance moved

R=u_x\times t

R=10\times 2.11

R=21.18 m

6 0
3 years ago
In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
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