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Fed [463]
3 years ago
10

Instead, suppose that it was fired upward at 60◦ with respect to a horizontal line. Then its horizontal component of velocity is

50 m/s. What would be the speed of the projectile at the top of its trajectory?
Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

50 m/s

Explanation:

Angle = 60 degree

Horizontal component of velocity = 50 m/s

A projectile motion is the motion of an object in two dimensions under the influence of gravity.

In this case, the object has no acceleration along horizontal direction, it has acceleration in vertical direction which is equal to the acceleration due to gravity of earth.

When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.

Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.

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A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one
allochka39001 [22]

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, mg, downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

T-mg=m\frac{v^2}{r}

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

mg=100 N is the weight of the ball

m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N

7 0
3 years ago
In which situation is maximum work considered to be done by a force
Katena32 [7]

Answer:

  • Work done is maximum when the movement of object is in line and direction of force.  

OR

  • Work done is maximum, when displacement takes place along the direction of force.
  • Work done is given by the equation

                                W = F.S

                            <em>     W  = F. S cos Θ</em>

<em>When cos Θ = 0° ; cos 0 = 1</em>

3 0
2 years ago
Read 2 more answers
A girl stand 5m away from a large plane mirror.How far must she walk ti be 2m away from her image?​
KonstantinChe [14]

Answer:

3m

Explanation:

If a girl is 5m away from a mirror, she needs to walk a further 3m to be 2m away from her image.

5m-3m = 2m

4 0
1 year ago
A wave is a:
liraira [26]

Answer:

wave

Explanation:

5 0
3 years ago
A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the
erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

Air temperature, T_{infty} = 20^{\circ}

k = 388 W/mK

h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

where

m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W

5 0
3 years ago
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