Ask question

Ask question

All categories

soldi70 [24.7K]

3 years ago

14

While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude

of the change in momentum of the ball?

1 answer:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

5.94 N·s

Explanation:

F = 132 N

t = 0.045 s

Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s

You might be interested in

HELP PLS MARKING BRANLIST 100 Pts TAKING TEST RN

vredina [299]

Answer:

0.5 m/s2

Explanation:

accelration formula : final velocty - starting velocity divided by time

5 0

2 years ago

How many centimeters are in 3.50 feet?

Kazeer [188]

106.68 centimetres are in 3.50 feet

4 0

3 years ago

For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.

Lelechka [254]

Answer:

B. 161.5 J

Explanation:

$n$ = Number of moles = 0.37

$\Delta T$ = Rise in the temperature of the oxygen gas = 15 K

$Q$ = heat added in order to raise the temperature

$c_{p}$ = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

$Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J$

6 0

3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

1 year ago

A well blended mixture that contains at least one solute and one solvent is a _____.

tensa zangetsu [6.8K]

A solution, an example is salt and water. brainliest?

4 0

3 years ago