What is the relationship between gravity and electromagatism?

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