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3 years ago

calculate the momentum of 5.07 kg egg that is dropped from a roof and falls 2 seconds before hitting the ground.

1 answer:

8 0

Anything that's dropped on or near the Earth accelerates at the rate of
9.8 m/s² on its way down. The mass of the object makes no difference.
After falling for 2 seconds, it's falling at the rate of 19.6 m/s.

Momentum = (mass) x (speed)

The egg's momentum is (5.07 kg) x (19.6 m/s) = 99.4 kg-m/s .

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A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landin

Kobotan [32]

<span>c) what is the angle (relative to the horizontal) of the balls initial velocity? </span>

8 0

3 years ago

Which of the following are true about a "simple compressible system"? It cannot be a mixture of different substances (e.g. oxyge

fomenos

Explanation:

The volume of a simple compressible system is not fixed. At a state of equilibrium, there should be uniformity in the entire system.

From the question we have here, these are the correct options:

1. It cannot be a mixture of different substances (e.g. oxygen and nitrogent)

2. It can be composed of any phases of a substance: solid, liquid, and/or gas

3. It's state is specified if given two independent, intensive thermodynamic properties.

4 0

3 years ago

In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c

shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

$P = \frac{F}{A}$

Hydraulic Lift - After change

$P + \Delta P = \frac{F + \Delta F}{A}$

Where:

$P$ - Hydrostatic pressure, measured in pascals.

$\Delta P$ - Change in hydrostatic pressure, measured in pascals.

$A$ - Cross sectional area of the hydraulic lift, measured in square meters.

$F$ - Hydrostatic force, measured in newtons.

$\Delta F$ - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

$\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}$

$\Delta P = \frac{\Delta F}{A}$

$\Delta F = A\cdot \Delta P$

Given that $\Delta P = 100\,Pa$ and $A = 25\,m^{2}$ , the additional weight is:

$\Delta F = (25\,m^{2})\cdot (100\,Pa)$

$\Delta F = 2500\,N$

The additional mass needed for the additional weight is:

$\Delta m = \frac{\Delta F}{g}$

Where:

$\Delta F$ - Additional weight, measured in newtons.

$\Delta m$ - Additional mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

If $\Delta F = 2500\,N$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }$

$\Delta m = 254.92\,kg$

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0

3 years ago

A 2.0-cm-diameter, 15-cm-long solenoid is tightly wound with one layer of wire. A 2.5 A current through the wire generates a 3.0

sweet [91]

Answer:

d = 1.047 mm

Explanation:

given,

diameter of the wire = 2.0-cm

length of solenoid = 15 cm = 0.15

Current in the wire = I = 2.5 A

magnetic field = B = 3.0 mT

Magnetic field inside the solenoid

$B = \dfrac{\mu_0 N I}{L}$

N x d = l

$N = \dfrac{l}{d}$

$B = \dfrac{\mu_0 \dfrac{l}{d} I}{L}$

$3 \times 10^{-3} = \dfrac{4\pi \times 10^{-7}\times \dfrac{0.15}{d}\times 2.5}{0.15}$

$0.45 \times 10^{-3}\ d = 4\pi \times 10^{-7}\times 0.15\times 2.5$

$\ d = \dfrac{4\pi \times 10^{-7}\times 0.15\times 2.5}{0.45 \times 10^{-3}}$

d = 1.047 x 10⁻³ m

d = 1.047 mm

diameter of the wire is d = 1.047 mm

5 0

3 years ago