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4 years ago

After a ball is thrown upward and is in the air, what will its acceleration do?

Physics

1 answer:

kifflom [539]4 years ago

8 0

Acceleration will always be samei.e.

Acceleration=g= 9.81 m/s^2

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m_a_m_a [10]

JAYLA OTHER KIDS ANSWER OUR QUESTIONS DID YOU KNOW THAT BESTFRIEND

4 0

4 years ago

How can I connect projectile motion into inclines. plsss give me ideas

Alenkasestr [34]

Answer:

Explanation:

Initial launch angle, θ

Initial velocity, u.

Time of flight, T.

Acceleration, a.

Horizontal velocity, vx.

Vertical velocity, vy.

Displacement, d.

Maximum height, H.

4 0

4 years ago

A 20 g ball of clay traveling east at 4.5 m/s collides with a 45 g ball of clay traveling north at 2.0 m/s. You may want to revi

german

Answer:

The ball travels with a speed of 1.96 m/s in a North East direction

Explanation:

Based on the law of conservation of linear momentum, we have that

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Since this problem involves motion in both the x and y coordinates, we will solve it in the separate coordinates, and then find the resultant as our answer.

Momentum in the x- direction

in the x direction, $v_{2}= 0$ . this makes $m_{2}v_{2}=0$

hence we have

$20\times4.5=65v_{x}$

$v_{x}=1.385m/s$

Momentum in the y- direction

in the y direction, $v_{1}= 0$ . this makes $m_{1}v_{1}=0$

hence we have

$45\times2.0=65v_{y}$

$v_{y}=1.385m/s$

The resultant of the two vectors can be found using Pythagoras' theorem

$v=\sqrt{1.385^{2}+1.385^{2}} =1.959$

The direction of the resultant vector is found as

$Tan^{-1}(\frac{1.385}{1.385})=45degrees$

Hence the ball moves with a velocity of 1.96m/s in a North East direction

7 0

3 years ago

The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/

svp [43]

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is $\frac{1}{6}$

So, the equation is

$\frac{1}{6} = 1 - \frac{T_2}{T_1}$

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is $\frac{1}{3}$

Therefore the next equation is

$\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}$

Now solve both the equations

solve equations (1) and (2)

$2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)$

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

7 0

4 years ago

A particle is located on the x axis 4.9 m from the origin. A force of 38 N, directed 30° above the x axis in the x-y plane, acts

Juli2301 [7.4K]

Answer:

Torque is 93 Nm anticlockwise.

Explanation:

We have value of torque is cross product of position vector and force vector.

A force of 38 N, directed 30° above the x axis in the x-y plane.

Force, F = 38 cos 30 i + 38 sin 30 j = 32.91 i + 19 j

A particle is located on the x axis 4.9 m and we have to find torque about the origin on the particle.

Position vector, r = 4.9 i

Torque, T = r x F = 4.9 i x (32.91 i + 19 j) = 4.9 x 19 k = 93.1 k Nm

So Torque is 93 Nm anticlockwise.

7 0

3 years ago