Two tuning forks, 254 Hz. and 260 Hz., are struck simultaneously. How many beats will be heard?

Tony ran 600 meters in 60 seconds. What was Tony's speed during the race?

NARA [144]

10 meters per second.

4 0

3 years ago

Read 2 more answers

Can someone please give me the answers to this? ... please ...

alexira [117]

3. 2 meters per second 4. i think object 7. i’ll try to figure it out 8. .77 9. 25km 10. 10m each second

7 0

3 years ago

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa

Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

$\frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

a ) when charge remains constant

energy = $\frac{q^2}{2C}$

q is charge and C is capacity

energy stored initially E₁= $\frac{q^2}{2C_1}$

energy stored finally E₂ = $\frac{q^2}{2C_2}$

$\frac{E_1}{E_2} = \frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

$E_2$ = $\frac{1.15}{2.3 } \times E_1$

= $\frac{1.15}{2.3 } \times 8.38$

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

$\frac{E_2}{E_1} = \frac{C_2}{C_1}$

$\frac{E_2}{8.38} = \frac{2.3}{1.15}$

$E_2$ = 16.76 .

8 0

3 years ago

A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

3 years ago

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$