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schepotkina [342]

3 years ago

14

A ball whose mass is 0.3 kg hits the floor with a speed of 5 m/s and rebounds upward with a speed of 2 m/s. If the ball was in c

ontact with the floor for 1.5 ms (1.5multiply10-3 s), what was the average magnitude of the force exerted on the ball by the floor?

1 answer:

Sonbull [250]3 years ago

4 0

Answer:

1400 N

Explanation:

Change in momentum equals impulse which is a product of force and time

Change in momentum is given by m(v-u)

Equating this to impulse formula then

m(v-u)=Ft

Making F the subject of the formula then

$F=\frac {m(v-u)}{t}$

Take upward direction as positive then downwards is negative

Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

$F=\frac {0.3(2--5)}{0.0015}=1400N$

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Which of the following is a type of T cell?

Bess [88]

Answer:

D

Explanation:

7 0

3 years ago

Read 2 more answers

A red blood cell is 7 x 10-6 m in size. This means it is 7 __________ in size. What unit completes the sentence?

aalyn [17]

If a red blood cell is 7 x $10^-^6$ m in size, it means that such a cell is 7 microns or micrometer in size. The unit that completes the sentence will, thus, be microns or micrometer.

According to international system of measurement:

$10^{-1}$ m = decimeter

$10^{-2}$ m = centimeter

$10^{-3}$ m = millimeter
$10^{-6}$ m = micrometer or micron
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Thus, a measurement of 7 x $10^-^6$ m is the same is 7 micons or 7 micrometer.

More on metric conversion can be found here: brainly.com/question/17767575

5 0

2 years ago

astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o

devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ $T=2 \pi\sqrt{\frac{m}{k} }$

or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

⇒ $=\frac{280\times (3)^2}{4 \pi^2}$

⇒ $=\frac{280\times 9}{4\times (3.14)^2}$

⇒ $=68.83 \ kg$

(b)

The speed of the string will be:

⇒ $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒ $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒ $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒ $=\sqrt{\frac{280\times 0.12}{63.83} }$

⇒ $=0.725 \ m/s$

4 0

3 years ago

e-lub [12.9K]

The engineer built a device called a generator

4 0

3 years ago

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

5 0

3 years ago