How the properties of sound waves change as they spread out in a spherical pattern

A tennis ball (m=0.060 kg) is moving horizontally at 20 m/s toward a tennis player who hits it straight back at 26 m/s. What is

Rus_ich [418]

Answer:

0.36 kg-m/s

Explanation:

Given that,

Mass of a ball, m = 0.06 kg

Initial velocity of the ball, u = 20 m/s

Final velocity of the ball, v = 26 m/s

We need to find the change in momentum of the tennis ball. It is equal to the final momentum minus initial momentum

$\Delta p=m(v-u)\\\\=0.06\times (26-20)\\\\=0.36\ kg-m/s$

So, the change in momentum of the ball is 0.36 kg-m/s.

4 0

2 years ago

5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-

Airida [17]

D. 980, this is the best answer because 35 x 7 is 980 :)

3 0

2 years ago

If a car is moving to the left with constantvelocity, one can conclude that

jeka94

Answer:

The net force applied to the car is zero.

Explanation:

We are given that a car is moving to the left with constant velocity.

When the car moving with constant velocity

Then, the final velocity=Initial velocity

Change in velocity=Final velocity- initial velocity=0

When change in velocity is zero then , acceleration of car

$a=\frac{change\;in\;velocity}{time}=\frac{0}{t}=0$

When acceleration is zero then, By Newtons second law

$Force=Mass\times acceleration=Mass\times 0=0$

The net force applied on the car will be zero.

Option C:The net force applied to the car is zero.

5 0

3 years ago

Daniel rode his bike 35km in 6 hours. What was his speed?

spayn [35]

His speed per hour was 5.8km and the total of 6 hours is 35km

8 0

3 years ago

Read 2 more answers

The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls l

Andreas93 [3]

Let $a$ be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

$1.3mg - T = 1.3ma$

where $T$ is the magnitude of tension in the pulley cord, and

• for the mass on the right,

$T - mg = ma$

Eliminate $T$ to get

$(1.3mg - T) + (T - mg) = 1.3ma + ma$

$0.3mg = 2.3ma$

$\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}$

Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity $v$ such that

$v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}$

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

$mg - T' = ma'$

where $T'$ and $a'$ are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

$T' - 0.8mg = 0.8ma'$

Eliminate $T'$ .

$(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'$

$0.2mg = 1.8 ma'$

$\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}$

Now, the right-hand mass has an initial upward velocity of $v$ , but we're now treating down as the positive direction. As it returns to its starting position, its speed $v'$ at that point is such that

${v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}$

3 0

1 year ago