All categories

3 years ago

To show Rmax=4Hmax, where Hmax is maximum height attained and Rmax is maximum range of a projectile

Physics

1 answer:

ladessa [460]3 years ago

7 0

Rmax = 4Hmax, that happen if the elevation of a projectile is 45°

You might be interested in

A 89.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.77 r

balu736 [363]

Answer:

$\omega_{2} = 1.107\,\frac{rad}{s}$

Explanation:

Let assume that circular platform is a solid cylinder. Given the absence of external forces, the situation can be analyzed by applying the Principle of Angular Momentum, which states that:

$I_{1}\cdot \omega_{1} = I_{2}\cdot \omega_{2}$

The initial moment of inertia is:

$I_{1} = \frac{1}{2}\cdot (89.3\,kg)\cdot (1.69\,m)^{2}$

$I_{1} = 127.525\,kg\cdot m^{2}$

Likewise, the final moment of inertia is:

$I_{2} = 127.525\,kg\cdot m^{2} + (8.83\,kg)\cdot (1.352\,m)^{2} + (21.1\,kg)\cdot (1.69\,m)^{2}$

$I_{2} = 203.929\,kg\cdot m^{2}$

The final angular speed is:

$\omega_{2} = \frac{I_{1}}{I_{2}} \cdot \omega_{1}$

$\omega_{2} = \frac{127.525\,kg\cdot m^{2}}{203.929\,kg\cdot m^{2}} \cdot (1.77\,\frac{rad}{s} )$

$\omega_{2} = 1.107\,\frac{rad}{s}$

4 0

3 years ago

Read 2 more answers

Which of the following is a pure substance that can't be separated by physical means, and can't be decomposed into other substan

g100num [7]

Your answer is C. element

6 0

3 years ago

Read 2 more answers

What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms

Alexus [3.1K]

Answer:

1.5F

Explanation:

Using

E= F/q

Where F= force

E= electric field

q=charge

F= Eq

So if qis tripled and E is halved we have

F= (E/2)3q

F= 1.5Eq=>> 1.5F

4 0

3 years ago

A train going 14m/s moves 250 m while accelerating to a stop. What is the train’s deceleration?

Elanso [62]

Answer:

-0.056 is the deceleration

6 0

3 years ago

Can y’all please help me with this 3 part question?

klio [65]

Answer:

Vf = 210 [m/s]

Av = 105 [m/s]

y = 2205 [m]

Explanation:

To solve this problem we must use the following formula of kinematics.

$v_{f} =v_{o} +g*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (released from the rest)

g = gravity acceleration = 10 [m/s²]

t = time = 21 [s]

Vf = 0 + (10*21)

Vf = 210 [m/s]

Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)

The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)

Av = (210 + 0)/2

Av = 105 [m/s]

To calculate the distance we must use the following equation of kinematics

$v_{f} ^{2} =v_{o} ^{2} +2*g*y\\\\(210)^{2} = 0 + (2*10*y)$

44100 = 20*y

y = 2205 [m]

5 0

3 years ago