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2 years ago

If a car is moving at a constant velocity of 10 m/s, what is its acceleration?

Physics

1 answer:

marin [14]2 years ago

4 0

Answer:

If a car is moving at a <em><u>constant velocity of 10 m/s,</u></em> there will be no change in the velocity per time.

Acceleration is the rate of change of velocity

Acceleration=(Final velocity-Initial velocity)/Time

Acceleration=(10m/s-10m/s)/Time= <em><u>0 m/s²</u></em>

<h3>★ <u>0 m/s²</u> is the right answer. </h3>

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Use Snell's Law to solve the following:

azamat

Answer:

1.171

Explanation:

if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;

$n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705$

3 0

3 years ago

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 dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

Olegator [25]

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

6 0

2 years ago

Read 2 more answers

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

3 years ago

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Juliette [100K]

<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, <u>as long as the spring is not permanently deformed</u>: