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3 years ago

If there is a 25 [lbs] of force acting on a 5 [lbs] of mass, what is the acceleration of that mass?

Physics

1 answer:

mihalych1998 [28]3 years ago

5 0

Answer:

f = 25 lbs

m = 5 lbs

a =?

f = ma

25 = 5 a( divide both sides by 5)

a = 5(lbs)

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A spaceship accelerates from 0m/s to 40m/s in 5 seconds. What is the acceleration of the spaceship

nadya68 [22]

Acceleration = change in velocity/time
= 40/5
=8m/s^2

6 0

3 years ago

A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro

prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope= $.2\frac{kg}{m}$

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as, $W_{1}=Force \times displacement$

= $\int\limits^{15}_0 {.2x} \, dx$ ..............(1)

= $.1\times 15^2$

=22.5 Nm

work done in lifting the sand is given as, $W_{2}=Force \times displacement$

$W_{2}=\int\limits^{15}_0 F \, dx$ .................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m= $\frac{20-18}{15-0}$

m=.133

Therefore,

$F=.133x+2$

Put value of F in equation 2

$W_{2}=\int\limits^{15}_0 (.133x+2) \, dx$

$W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm$

Therefore,

work done lifting the bucket (sand and rope) to the top of the building, $W=W_{1}+W_{2}$

W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0

3 years ago

How many SF does number 0.00403 have?

snow_tiger [21]

The answer is two (4 -3)
Zero is not a SF

6 0

3 years ago

Need help with these please

Montano1993 [528]

Answer:

Explanation:

Which number all or one?

6 0

3 years ago

Read 2 more answers

A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(

Flura [38]

Answer:

8.97 Watt

Explanation:

Resistance, R = 20 ohm

Inductance, L = 10 mH

V(t) = 20 Cos (1000 t + 45°)

Compare with the standard equation

V(t) = Vo Cos(ωt + Ф)

Ф = 45°

ω = 1000 rad/s

Vo = 20 V

Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm

impedance is Z.

$Z = \sqrt{R^{2}+X_{L}^{2}}$

$Z = \sqrt{20^{2}+10^{2}}$

Z = 22.36 ohm

$V_{rms}=\frac{V_{0}}{\sqrt{2}}$

$V_{rms}=\frac{20}{\sqrt{2}} = 14.144 V$

$I_{rms}=\frac{V_{rms}}{Z}=\frac{14.144}{\sqrt{22.36}}=0.634 A$

Apparent power is given by

P = Vrms x Irms

P = 14.144 x 0.634

P = 8.97 Watt

6 0

3 years ago