An alpha particle is identical to a:

A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

LekaFEV [45]

Answer:

Height is 11.25m

Explanation:

Given the following data;

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

$V^{2} = U^{2} + 2aS$

Making S the subject, we have;

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

Substituting into the equation, we have;

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

Therefore, the ball will reach a height of 11.25m before it begins to fall.

4 0

2 years ago

A 1200 kg car traveling north at 10 m/s is rear-ended by a 2000 kg truck traveling at 30 m/s. What is the total momentum before

tino4ka555 [31]

Answer:

The total momentum before and after collision is 72000 kg-m/s.

Explanation:

Given that,

Mass of car = 1200 kg

Velocity of car = 10 m/s

Mass of truck = 2000 kg

Velocity of truck = 30 m/s

Using conservation of momentum

The total momentum before the collision is equal to the total momentum after collision.

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

Where, $m_{1}$ =mass of car

$v_{1}$ =velocity of car

$m_{1}$ =mass of truck

$v_{1}$ =velocity of truck

Put the value into the formula

$1200\times10+2000\times30=(1200+2000)V$

$V=\dfrac{1200\times10+2000\times30}{(1200+2000)}$

$V = 22.5\ m/s$

Now, The total momentum before collision is

$P=m_{1}v_{1}+m_{2}v_{2}$

$P=1200\times10+2000\times30$

$P=72000\ kg-m/s$

The total momentum after collision is

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

$P=(1200+2000)\times22.5$

$P= 72000 kg-m/s$

Hence, The total momentum before and after collision is 72000 kg-m/s.

4 0

2 years ago

How do inertia and centripetal force combine to keep an object moving in circular motion?

katen-ka-za [31]

As the centripetal force acts upon an object moving in a circle at constant speed, the force always acts inward as the velocity of the object is directed tangent to the circle. ... In fact, whenever the unbalanced centripetal force acts perpendicular to the direction of motion, the speed of the object will remain constant.

6 0

2 years ago

Read 2 more answers

I'm spending 98 points please help!

juin [17]

Answer: C. 8.0 m west

Explanation: The arrows are going 15 m west and 7.0 m east. 7 meters of the west will cancel out because 15-7=8. Subtract the smaller number from the bigger number, which is west minus east. The answer will be 8.0 m west.

5 0

3 years ago

Read 2 more answers

a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s

Kobotan [32]

Solving this using the time, we know that range = horizontal velocity x time of flight

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have

range = 36 cos 28 x 3.44 s = 109.3 m

6 0

2 years ago