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2 years ago

an object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of

1 answer:

5 0

An object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of θ with the horizontal.

1. Which direction does the force of gravity act on the object?

2. Which direction does the normal force act on the object?

Which force is responsible for the object moving down the incline?

Answer:

The answer is below

Explanation:

1. When an object is moving with a constant velocity, the direction the force of gravity act on the object is DIRECTLY DOWN.

2. When an object is moving with a constant velocity, the direction the normal force act on the object "perpendicular to the surface of the plane."

3. When an object is moving with a constant velocity, the force that is responsible for the object moving down the incline is "the component of the gravitational force parallel to the surface of the inclined plane."

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QUESTION 1

mario62 [17]

It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.

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2 years ago

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A house is losing heat at a rate of 1600 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

Setler [38]

Answer:

1600 kJ/h per K, 888.88 kJ/h per °F and 888.88kJ/h per R

Explanation:

We make use of relations between temperature scales with respect to degrees celsius:

$1 K= 1^{\circ}C+273\\1^{\circ}F= (1^{\circ}C*1.8)+32\\1 R= (1^{\circ}C*1.8)+491.67$

This means that a change in one degree celsius is equivalent to a change of one kelvin, while for a degree farenheit and rankine this is equivalent to a change of 1.8 on both scales.

So:

$\frac{Q}{\Delta T(K)}=\frac{Q}{\Delta T(^\circ C)}=1600 \frac{kJ}{h} per K\\\frac{Q}{\Delta T(^\circ F)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per ^\circ F\\\frac{Q}{\Delta T(R)}=\frac{Q}{\Delta T(^\circ C*1.8)}=888.88 \frac{kJ}{h} per R$

5 0

2 years ago

When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is incre

Crazy boy [7]

Answer:

the previous correct answer is b

Explanation:

When the circuit is closed in the system, a current is induced that follows the lenz law, which opposes the change that is occurring and therefore the coil increases and the idicidal current in the ring must reach the maximum oppositing is the current of the coil, so quiet force is repulsion

Consequently, the previous correct answer is b

3 0

2 years ago

. A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.20 104 kg. To dive, this submarine takes on

dangina [55]

Answer:

Explanation:

Given that,

Bathysphere radius

r = 1.5m

Mass of bathysphere

M = 1.2 × 10⁴ kg

Constant speed of descending.

v = 1.2m/s

Resistive force

Fr = 1100N upward direction

Density of water

ρ = 1.03 × 10³kg/m³

The volume of the bathysphere can be calculated using

V = 4πr³ / 3

V = 4π × 1.5³ / 3

V = 14.14 m³

The Bouyant force can be calculated using

Fb = ρgV

Fb = 1.03 × 10³ × 9.81 × 14.14

Fb = 142,846.18 N

Buoyant force is acting upward

Weight of the bathysphere

W = mg

W = 1.2 × 10⁴ × 9.81

W = 117,720 N

Weight is acting downward

The net positive buoyant using resolving

Fb+ = Fb — W

Fb+ = 142,846.18 — 117,720

Fb+ = 25,126.18 N

The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force

W = Fb+ + Fr

W = 25,126.18 + 1100

W = 26,226.18

mg = 26,226.18

m = 26,226.18 / 9.81

m = 2673.4kg

Mass of submarine is 2673.4kg

8 0

2 years ago

Two coils, held in fixed positions, have a mutual inductance of M = 0.0034 H. The current in the first coil is I(t) = I0sin(ωt),

tiny-mole [99]

Answer:

ε₂ =2.63 V

Explanation:

given,

M = 0.0034 H

I (t) = I₀ sin (ωt)

I (t) = 5.4 sin (143 t)

$\dfrac{d i(t)}{dt} = \dfrac{d}{dt}(5.4 sin (143 t))$

$\dfrac{d i(t)}{dt} =143 \times 5.4 cos (143 t)$

magnitude of the induced emf in the second coil

ε₂ = $M\dfrac{di}{dt}$

ε₂ = $0.0034\times 143 \times 5.4 cos (143 t)$

for maximum emf

cos (143 t) = 1

ε₂ = $0.0034\times 143 \times 5.4$

ε₂ =2.63 V

3 0

2 years ago