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3 years ago

When a ball bounces back when you throw it at a wall, what occurs?

Physics

2 answers:

NISA [10]3 years ago

5 0

Answer:

a (refraction) i think

Explanation:

Ray Of Light [21]3 years ago

4 0

I would have to think a. Nothing else makes since

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Ibrahim is experimenting with his circuits. He made a circuit using a 12V battery and 6.5 ohm resistor. But then he realized he

SVETLANKA909090 [29]

7.5 V im pretty sure.

4 0

3 years ago

Read 2 more answers

"2.40 A pressure of 4 × 106N/m2 is applied to a body of water that initially filled a 4300 cm3 volume. Estimate its volume after

wel

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as

K =-V dP/dV

Where K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV

dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

3 0

3 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago

To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If

arlik [135]

Answer:

Part a)

$a = 1260.3 m/s^2$

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

$v_f^2 - v_i^2 = 2 a d$

$v_1^2 - 0^2 = 2(9.81)(4.0)$

$v_1 = 8.86 m/s$

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - v_2^2 = 2(-9.81)(2.00)$

$v_2 = 6.26 m/s$

Part a)

Average acceleration is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}$

$a = 1260.35 m/s^2$

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0

3 years ago

Most earthquakes occur along or near the edges of the earth's

Pepsi [2]

Most earthquakes occur along or near the edges of the earth's lithospheric<span> plate. </span>

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4 years ago