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mojhsa [17]
3 years ago
9

A 1500 kg truck is acted upon by a force that decreases its speed from 25 m/s to 15 m/s in 8 s. What is the magnitude of the for

ce acting on the truck?
Physics
1 answer:
Citrus2011 [14]3 years ago
4 0

Answer:

F = 1,875 N

Explanation:

force=

\frac{change \: in \: momentum}{time \: taken}

∆H = m∆V

where ∆H ----> change in momentum.

( final momentum - initial momentum )

and ∆V ----> change in velocity

( final velocity - initial velocity )

and m ----> is mass

then f =

\frac{1500 \times (25 - 15)}{8}

= 1,875 N

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Tanya [424]
c) 101kPa

Hope I helped! ( Smiles )
5 0
3 years ago
Read 2 more answers
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
Letters A B, C and D represent locations on a bar magnet. Which location has the greatest magnetic force? (l) A (3) C ets (2) B
Elena L [17]
Answer: The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center. So it would be D.

6 0
3 years ago
a student measure the mass of an 8 cm^3 block of brown sugar to be 12.9 g. what is the density of the brown sugar?
schepotkina [342]
Density = mass / volume ;
1 Cubic Centimeter = 0,000001 Cubic Meter
8 cm^3 = 0.000008 m^3
12,9 g = 0,0129 kg


The density is 0,0129 kg/ 0,000008 m^3 = <span><u>1612,5 kg/m^3</u> </span>
4 0
4 years ago
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