01:28:29

A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi

Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0

3 years ago

Do people use uranium for anything?

kati45 [8]

Answer:

kjlp;;

Explanation:

7 0

3 years ago

Read 2 more answers

Calculate the number of moles of water molecules in 12 dm' of water vapour at STP.

Vinvika [58]

Answer:

$22.4 \: {dm}^{3} \: are \: occupied \: by \: 1 \: mole \\ 12 \: {dm}^{3} \: will \: be \: occupied \: by \: ( \frac{12}{22.4} ) \: moles \\ = 0.536 \: moles$

4 0

3 years ago

A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between

Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

$a = \dfrac{F}{m}$

Where, F = frictional force

m = mass

Put the value into the formula

$a=\dfrac{\mu mg}{m}$

$a=\mu g$

$a=0.229\times9.8$

$a=2.244\ m/s^2$

We need to calculate the speed of the sled

Using equation of motion

$v^2=u^2-2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v ^2=(5.49)^2-2\times2.244\times3.89$

$v=\sqrt{(5.49)^2-2\times2.244\times3.89}$

$v=3.56\ m/s$

Hence, The speed of the sled is 3.56 m/s.

8 0

3 years ago

24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is

Wewaii [24]

Given :-

An elevator is moving vertically up with an acceleration a.

To Find :-

The force exerted on the floor by a passenger of mass m .

Solution :-

As the man is in a accelerated frame that is non inertial frame , we would have to think of a pseudo force .

The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

$\implies Weight_{apparent}= mg + ma$

Hence option d is correct choice .

I hope this helps .

3 0

3 years ago