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3 years ago

A ball has mass of 140g. what is the force to accelerate the ball at 25m/s

Physics

2 answers:

adelina 88 [10]3 years ago

8 0

Hello there!

formula: F=ma

It would be 3.5 N

Fittoniya [83]3 years ago

6 0

We know, F = m * a

Here, m = 140 g = 0.140 Kg

a = 25 m/s2

It would be: F = 0.140 * 25 = 3.5 N

So your answer would be 3.5N

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3. A satellite is in orbit around some mystery planet. You observe that it takes 3 earth days, or (86,400*3) seconds for this sa

Iteru [2.4K]

Answer:

3 x 10^18 kg

Explanation:

Time period, T = 3 days = 86400 x 3 = 259200 seconds

r = 7 x 10^5 m

Let M be the mass of planet

Use the formula of time period of satellite

$T = 2\pi \sqrt{\frac{r}{GM}}$

Where, G be the universal gravitational constant.

$M=\frac{4\pi ^{2}r^{3}}{GT^{2}}$

By substituting the values

$M=\frac{4\times 3.14 \times 3.14\times \left ( 7\times 10^{5} \right )^{3}}{6.67\times 10^{-11}\times 259200\times 259200}$

M = 3 x 10^18 kg

Thus , the mass of planet is 3 x 10^18 kg.

3 0

3 years ago

A vehicle approaches a set of traffic lights with a speed of 15m/s. When 50m from the lights the driver brakes and comes to a re

sattari [20]

Answer:

–2.25 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 15 m/s

Distance travelled (s) = 50 m

Final velocity (v) = 0 m/s

Deceleration (a) =?

v² = u² + 2as

0² = 15² + (2 × a × 50)

0 = 225 + 100a

Collect like terms

0 – 225 = 100a

– 225 = 100a

Divide both side by 100

a = –225/100

a = –2.25 m/s²

Thus, the deceleration of the vehicle is –2.25 m/s²

3 0

3 years ago

As it rains and snows near mountaintops, the water is transferred from the atmosphere back onto ground. What natural force pulls

Semenov [28]

Gravity pulls the water down off the mountain, witch is run-off.

8 0

4 years ago

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which

emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

= .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s

8 0

3 years ago

An ideal gas initially at 4.00atm and 350 K is permitted

Nuetrik [128]

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is $P_{2}$ and final temperature is $T_{2}$ .

(a) We know that for a monoatomic gas, value of $\gamma$ is \frac{5}{3}[/tex].

And, in case of adiabatic process,

$PV^{\gamma}$ = constant

also, PV = nRT

So, here $T_{1}$ = 350 K, $V_{1} = V$ , and $V_{2} = 1.5 V$

Hence, $\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}$

$\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}$

$T_{2}$ = 267 K

Also, $P_{1}$ = 4.0 atm, $V_{1} = V$ , and $V_{2} = 1.5 V$

$\frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}$

$\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}$

$P_{2}$ = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of $\gamma$ is \frac{7}{5}[/tex].

As, $PV^{\gamma}$ = constant

also, PV = nRT

$T_{1}$ = 350 K, $V_{1} = V$ , and $V_{2} = 1.5 V$

$\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}$

$\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}$

$T_{2}$ = 289 K

And, $P_{1}$ = 4.0 atm, $V_{1} = V$ , and $V_{2} = 1.5 V$

$\frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}$

$\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}$

$P_{2}$ = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.

6 0

3 years ago