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3 years ago

The broom hocket fame helps you work on what skill?

Physics

1 answer:

Solnce55 [7]3 years ago

7 0

(B) "Hand/foot coordination"

and if you don't mind me asking what is this for? lol what subject

You might be interested in

An object, initially at rest moves 250m in 17s. What is it's acceleration?

Gekata [30.6K]

With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

d= 250s
t= 17s

a= d/t
= $\frac{250}{17}$
= 4.7 $\frac{m}{s}$

8 0

2 years ago

When you approach a yield sign, while trying to enter or merge onto another road, traffic already on that road has the right of

kogti [31]

TRUE. When you approach a yield sign, while trying to enter or merge onto another road, traffic already on that road has the right of way.

5 0

2 years ago

Read 2 more answers

A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.

nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

<em>momentum of the ball, P = 0.9 kgm/s</em>
<em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

$Ft = \Delta P$

where;

$Ft$ is impulse

$\Delta P$ is change in momentum

The time of motion of the ball is calculated as follows;

$t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s$

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0

2 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$