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Brrunno [24]
3 years ago
7

Two forces are exerted on an object in the vertical direction: a 20 N force downward and a 10 N force upward. The mass of the ob

ject is 25 kg. (1) What are some possibilities about the motion of this object? (2) Represent the motion of the object with a force diagram and a motion diagram.
Physics
1 answer:
VMariaS [17]3 years ago
6 0

Answer:

They are equal.

Explanation:

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A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

So option (A) is correct option

Explanation:

Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

And velocity after rebound v_2=-19m/sec

We have to find the change in momentum

So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
3 years ago
Explain why the boiling point of a liquid varies with atmospheric pressure
OLEGan [10]
The boiling point of water, or any liquid, varies according to the surrounding atmospheric pressure. A liquid boils, or begins turning to vapor, when its internal vapor pressure equals the atmospheric pressure.
5 0
3 years ago
Select the correct description of how inductances are combined in series and in parallel
Ksivusya [100]

Answer:

it will be C

Explanation:

it will be C...

6 0
3 years ago
A jet plane flying 600 m/s experience an acceleration of 4.0 g when pulling out of a circular dive. What is the radius of curvat
aksik [14]

Answer:

Option C

Solution:

As per the question:

Velocity of the jet, v = 600 m/s

Acceleration, a = 4.0 g

Now,

To calculate the radius of curvature:

The necessary centripetal force on the jet is given by the force:

F = F_{c}                                               (1)

where

F_{c} = \frac{mv^{2}}{R} = centripetal force

where

m = mass of the jet

R = radius of curvature of the path

Using eqn (1):

ma = \frac{mv^{2}}{R}

Thus

a = \frac{v^{2}}{R}

4.0 g = \frac{600^{2}}{R}

where

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

4.0\times 9.8 = \frac{600^{2}}{R}

R = \frac{600^{2}}{4.0\times 9.8}

R = 9183.67 m ≈ 9200 m

3 0
3 years ago
A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved
MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s

Now we can solve the problem by applying the suvat equation:

v_y^2-u_y^2=2as

where

v_y is the vertical velocity when the pea hits the ceiling

a=g=-9.8 m/s^2 is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for v_y,

v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s

Therefore, the speed of the pea when it hits the ceiling is

v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s

5 0
3 years ago
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