Which coastline of South America is nearest to the oldest seafloor

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

HACTEHA [7]

Explanation :

Speed of car, $v_1=50\ km/h=13.8\ m/s$

Kinetic energy of the car, $KE_1=400\ J$

Speed of car, $v_2=100\ km/h=27.7\ m/s$

Let $KE_2$ is the kinetic energy of the car when it is moving with 100 km/h.

$\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}$

$\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}$

$KE_2=KE_1(\dfrac{v_2}{v_1})^2$

$KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2$

$KE_2=1611.6\ J$

Initial velocity of both cars are 0. Using third equation of motion :

So, $S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}$

and $S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}$

t is same. So,

$\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}$

$\dfrac{S_1}{S_2}=\dfrac{1}{2}$

$S_2=2\ S_1$

So, the braking distance at the faster speed is twice the braking distance at the slower speed.

4 0

4 years ago

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I don't understand why when you use a straw the higher pressure outside of it pushes the water up the straw.

ValentinkaMS [17]

When material stuff is free to move, it moves away from stronger force on it to weaker force on it. OK ? Now take it to the next step: When material stuff is free to flow ... like liquid or gas ... it flows from higher pressure to lower pressure. That's what makes wind, and that's why water comes out of a garden hose when you open the nozzle. It doesn't amaze you at all when you INCREASE the pressure at one end of the straw (blow into it), and air moves DOWN and out of it. So it shouldn't bother you that when you DECREASE the pressure at your end, anything that's free to flow UP the straw will do that.

3 0

3 years ago

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When sound waves are reflected back to the source, the reflected sound is called

sergey [27]

It’s an echo :))))))

6 0

3 years ago

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Need help with science pls help me!! 40 points

sp2606 [1]

Answer: a. Regularly, you rub your hands for warmth.

b. Some examples are friction from electricity going through a wire or friction from rubbing your hand together.

c. Without friction, we would fall every time we walked. Because there is friction, we can walk normally. Another example is a ball moving across the field. Without friction, the ball would not slow down.

Explanation:

Overall, friction is important to us because we depend on it.

3 0

2 years ago

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

3 years ago