How many neutrons are in an atom of carbon with a mass number of 13

Suppose Earth's gravitational force were decreased by half. How would this change affect a game of basketball? Write a paragraph

timama [110]

If the gravitational force were<span> decreased by half, there would be lack of gravity on earth. Hence, it would basically affect the velocity, speed, and the distance travelled in any direction by basketball players and the ball. The basketball would bounce higher and come down in a slower speed. Whereas for the players, they would be able to leap higher from the floor.</span><span> </span>

7 0

3 years ago

The driver sees that the road is empty and accelerates at 1.0 m/s2 for 5.0 s. what can you determine about the truck's motion us

vekshin1

kinematic equation

v=u+at

v-u=at

v-u = 1x5

the driver will have increased speed by 5 m/s. actual speeds unknown

7 0

3 years ago

A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a s

gtnhenbr [62]

Answer:

-387883.3 m/s²

0.000286168546055 seconds

Explanation:

t = Time taken

u = Initial velocity = 370 m/s

v = Final velocity = 259 m/s

s = Displacement = 9 cm

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2$

The acceleration of the bullet is -387883.3 m/s²

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s$

The time taken is 0.000286168546055 seconds

4 0

3 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

8 0

2 years ago

A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.

snow_tiger [21]

The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = $\sqrt{400/3.9} \approx 10.12739367$ , at = v, so $3.9 * 10.12739367 \approx 39.5 m/s$

6 0

3 years ago

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