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3 years ago

What is the acceleration of a 100 kg if a net force of 200 N is applied to it?

Physics

2 answers:

slamgirl [31]3 years ago

6 0

Answer: 2 m/s^2 or 6.562 ft/s^2

Zina [86]3 years ago

5 0

Your answer would be 2 m/s^2. I hope I could help you!

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If Michael Jordan has a vertical leap of 1.29 meters, what is his take-off speed and his hang time (total time to move upwards t

evablogger [386]

(Hint: the time to rise to the peakis one-half the total hang-time.).

6 0

4 years ago

Pressure of gas is 13.6 cm of hg what does it means

vlada-n [284]

Explanation:

13 cmHg (centimeters of mercury) is the pressure at the bottom of a column of mercury 13 cm deep. It is the equivalent of about 17.3 kPa or 2.5 psi.

5 0

3 years ago

A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a

uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

$m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}$

Where, $m_{h}$ = mass of halfback

$m_{c}$ =mass of corner back

$v_{h}$ = velocity of halfback

$v_{c}$ = velocity of corner back

Put the value into the formula

$98\times4.2+85\times5.5=(98+85)\times v$

$v=\dfrac{98\times4.2+85\times5.5}{98+85}$

$v=4.80\ m/s$

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0

3 years ago

Which of these is an example of acceleration ?

anygoal [31]

Answer:

no picture or anything so cant anwser

Explanation:

8 0

3 years ago

Read 2 more answers

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

4 years ago