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Step2247 [10]
3 years ago
13

What is the relationship between gravity and electromagatism?

Physics
1 answer:
Mashcka [7]3 years ago
5 0
Gravitational forece can only attract whereas electomagatism can be attractive and repulsive
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A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
Relative formula mass of CuCO3
Elodia [21]

AnMolar mass of CuCO3 = 123.5549 g/mol

This compound is also known as Copper(II) Carbonate.

Convert grams CuCO3 to moles  or  moles CuCO3 to grams

Molecular weight calculation:

63.546 + 12.0107 + 15.9994*3

Percent composition by element

Element   Symbol   Atomic Mass   # of Atoms   Mass Percent

Copper Cu 63.546 1 51.431%

Carbon C 12.0107 1 9.721%

Oxygen O 15.9994 3 38.848%

Explanation:

5 0
2 years ago
A 17 kg box sitting on a shelf has a potential energy of 350 J. How high is the shelf? Round your answer to the nearest whole nu
Free_Kalibri [48]
Potential energy is mass * gravity * height. (m*g*h).

350 = 17*9.8*h   <--350 is its energy, 17kg is its mass, and 9.8 is gravity's acceleration on the object. We now just need to solve for h.

h = 350/(17 * 9.8) =  2.1 meters, which, when rounded to the nearest whole meter, is 2 meters.

The shelf is 2 meters high.

3 0
3 years ago
Read 2 more answers
Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o
Andrews [41]

Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

8 0
3 years ago
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