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3 years ago

What’s a good hypothesis and and experiment for water?

Physics

1 answer:

Lelu [443]3 years ago

6 0

Answer: You could do something like, "how does water react to being mixed with baking soda"...or something along those lines

Explanation:

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A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw

irga5000 [103]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

6 0

3 years ago

022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

svetoff [14.1K]

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

Distance traveled by it in horizontal direction is given as

$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

6 0

2 years ago

Martha wants to calculate an object's velocity. What will she need to do?

rjkz [21]

Answer:

you divide the distance by the time it takes to travel that same distance, then you add your direction to it.

4 0

2 years ago

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound

Roman55 [17]

Answer:

a.96cm

b.77cm

Explanation:

Highest intensity, $d_1$ =19cm

Lowest intensity, $d_2$ =29cm

$d_1+d_2=19+29=48cm$

48cm is the distance for half-wave, wavelength =48cm*2=96cm

b. From a, above we know that half-wavelength is 48cm

Also, the distance of known maximum is 29cm

Next maximum is calculated as:

$Next \m\aximum=Last \ Maximum+ \ 0.5Wavelength\\=29cm+48cm\\=77cm$

The next maximum will occur at 77cm

6 0

3 years ago

At any angular speed, a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain unifo

Trava [24]

Answer:

d. 6/5 M.

Explanation:

let the angular velocity = $\omega$

the radius of sphere (r) = $\frac{D}{2}$

moment of inertia of solid sphere $I_s$ = $2/5 M(\frac{D}{2})^2$

rotational K.E of solid sphere= $\frac{1}{2}* I_s* \omega ^2$

= $\frac{1}{2} * 2/5 M(\frac{D}{2})^2 * \omega^2$

we represent the mass of hallow sphere= $M_{hs}$

moment of inertia of solid sphere $I_{ss}$ = $2/3 M_{hs}(\frac{D}{2})^2$

rotational K.E of hollow sphere= $\frac{1}{2}* I_{ss} * \omega^2$

= $1/2 *2/3 M_{hs}(\frac{D}{2})^2* \omega^2$

NOW,the kinetic energy of solid sphere= 1/2 kinetic energy of hallow sphere

= $1/2 [2/5 M(D/2)^2 ]\omega^2= 1/2 (1/2 (2/3 M_{hs}(D/2)^2 )\omega^2 }$

2/5M= 1/3 $M_{hs}$

$M_{hs}$ = 3(2/5M)

$M_{hs}$ = 6/5 M

7 0

3 years ago