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3 years ago

Which describes Einstein’s idea of space and time?

Physics

2 answers:

Sholpan [36]3 years ago

8 0

Time and space are both relative

max2010maxim [7]3 years ago

5 0

Time and space are both relative

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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

anzhelika [568]

Answer:

The position is $P = 47.4 \ m$ relative to the base of the ocean

Explanation:

From the question we are told that

The angle made by the incline with the horizontal is $\theta = 24.0 ^o$

The constant acceleration is $a = 3.82 \ m/s^2$

The distance covered is $d = 60.0 \ m$

The height of the cliff is $h = 50 .0 \ m$

The velocity of the car is mathematically represented as

$v^2 = u^2 + 2ad$

The initial velocity of the car is u= 0

$v^2 = 2ad$

substituting values

$v^2 = 2 * 3.82 * 60$

$v = 21.4 \ m/s$

The vertical component of this velocity is

$v_v = -v * sin(\theta )$

substituting values

$v_v = -21.4 * sin(24.0)$

$v_v = -8.7 \ m/s$

The negative sign is because is moving in the negative direction of the y-axis

The horizontal component of this velocity is

$v_h = v * cos (\theta)$

$v_h = 21.4 * cos (24.0)$

$v_h = 19.5 \ m/s$

Now according to equation of motion we have

$h = v_v*t - \frac{1}{2} * g t^2$

substituting values

$50 = -8.7 t - \frac{1}{2} * 9.8 t^2$

$4.9t^2 +8.7t -50 = 0$

using quadratic equation we have that

$t_1 = 2.42\ s \ and\ t_2 = -4.20\ s$

given that time cannot be negative

$t = 2.42 \ s$

The car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

$P = v_h * t$

substituting values

$P = 19.5 * 2.43$

$P = 47.4 \ m$

5 0

3 years ago

Looking that following diagram of bar magnets, determine if the magnets will or will not connect (attract) and why.

vovangra [49]

Answer: They will NOT connect because like poles are facing each other, and like poles repel each other.

3 0

4 years ago

A 0.03-kg bullet is fired with a horizontal velocity of 470 m/s and becomes embedded in block B which has a mass of 3 kg. After

Gala2k [10]

Answer with Explanation:

We are given that

Mass of bullet, $m_1=0.03 kg$

$u_1=470 m/s$

$m_2=3 kg$

$\mu_k=0.2$

$m_3=30 kg$

We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.

According to law of conservation of momentum

$m_1u_1=(m_1+m_2)v$

$0.03(470)=(0.03+3)v$

$v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s$

Hence, the velocity of the bullet and block B after the first impact=4.65 m/s

According to law of conservation of momentum

$(m_1+m_2)v=(m_1+m_2+m_3)V$

$(0.03+3)\times 4.65=(0.03+3+30)V$

$V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}$

$V=0.43 m/s$

3 0

4 years ago

Read 2 more answers

Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inner shell, which has a radius

Murrr4er [49]

Answer:

a) At 0.20 m, the magnitude of the field is 675.0 kV

The direction of the field is acting outwards from the center of the charged spheres

b) At 0.10 m, the magnitude of the field is 135 kV

The direction is acting outwards from the center of the charged spheres

c) At 0.025 m

The magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres

Explanation:

The charged spherical shell parameters are;

The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C

The radius of the inner shell, R₁ = 0.050 m

The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C

The radius of the outer shell, R₂ = 0.15 m

Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;

When r < R₁ < R₂,

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

When R₁ < r < R₂,

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

When R₁ < R₂ < r,

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

a) When r = 0.20 m, we have;

R₁ < R₂ < r, therefore

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} + 4.50\times 10^{-6} }{0.20^2} \right ) = 675.0 \ kV$

Therefore, the magnitude of the field, V = 675.0 kV

The direction of the field is outwards

b) When r = 0.10 m, we have;

When R₁ < r < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.10} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = 135 \ kV$

Therefore, the magnitude of the field, V = 135 kV

The direction of the field is outwards from the center

c) When r = 0.025 m, we have;

When r < R₁ < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.05} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = -270 \ kV$

Therefore, the magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres.

4 0

3 years ago

Which occurs when a wave is reflected?

dmitriy555 [2]

Answer: The wave can flip upside down.

Reflection is the bending of a wave when it cannot pass through. For example, plain mirrors which are flat, a ray of light hits the mirror and is reflected from the mirror since it cannot pass through

When reflection occurs the speed and frequency of the wave does not change but the wave is flipped upside down.

The speed does not change because speed is affected by the change in medium the frequency also remains the same since the energy of the wave does not change.

3 0

3 years ago

Read 2 more answers