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2 years ago

A projectile is thrown with an initial speed of 20 m/s at an angle 30 degrees above

1 answer:

4 0

Initial velocity u = 20 m/s

Initial horizontal velocity = 20 cos30° = 20 * 0.866 = 17.32 m/sec.

Initial vertical velocity = 20 sin30° = 20 * 1/2 = 10 m/sec.

time taken t = u/g = 10/10 =1 sec. ( approximating g to 10m/sec^2)

Maximum height h = ut + 1/2 * g * t^2

h = 10 *1 - 1/2 * 10 * 1* 1

h = 10 - 5 = 5 metres.

Total time in air = 2t = 2 seconds.

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What is the same about each type of electromagnetic wave?

xenn [34]

There all diffrent speeds

7 0

3 years ago

You observe that a mass suspended by a spring takes 0.25 s to make a full oscillation. What is the frequency of this oscillation

Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, $\omega = 25.13 rad/s$

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

f = $\frac{1}{T}$

f = $\frac{1}{0.25}$

Frequency of oscillation, f = 4 Hz

The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

Therefore, time period, T = 0.25 s

Angular frequency of oscillation is given by:

$\omega = 2\pi \times f$

$\omega = 2\pi \times 4$

$\omega = 25.13 rad/s$

5 0

3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

3 years ago

How much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate

Nuetrik [128]

Answer:

<h3>2479 Newton</h3>

Solution,

Mass=100 kg

Acceleration due to gravity(g)=24.79 m/s^2

Now,.

 $weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton$ 

hope this helps ..

Good luck on your assignment..

5 0

3 years ago

Space-shuttle astronauts experience accelerations of about 35 m/s2 during takeoff. What force does a 75 kg astronaut experience

amm1812

Answer:

<h3>The answer is 2625 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 75 × 35

We have the final answer as

Hope this helps you

5 0

3 years ago