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2 years ago

A fast-moving car has 0.5 MJ of kinetic energy.

Physics

1 answer:

DerKrebs [107]2 years ago

6 0

Answer:

-0.5 MW

Explanation:

For this we can use the principle of work and energy, which states that:

T₁ + ∑U₁₋₂ = T₂

Where T is kinetic energy (in situations one and two) and U is the sum of all work done (between situations one and two).

So, since the car is first moving with a kinetic energy of 0.5 MJ:

T₁ = 0.5 MJ = 0.5 · 10⁶ J

On top of this, we know that the car comes to a halt, using this as situation 2:

T₂ = 0 J

Filling in the equation:

⇒ 0.5 · 10⁶ + U = 0

⇒ U = -0.5· 10⁶ [W] (unit of work is watts)

= -0.5 MW

Thinking about the sign convention, we see that the brake force will point to the back of the car (backward), and the distance the car will travel before stopping will be forward, oposite directions and thus negative work.

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Elastic Energy is stored in it.

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2 years ago

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Find the net force and acceleration. 15 points. Will give brainliest!

gladu [14]

Answer:

$\boxed{F_{net} = 28.7 \ N}$

$\boxed{a = 2.1 \ m/s^2}$

Explanation:

Finding the net force:

Firstly , we'll find force of Friction:

$F_{k} = (micro)_{k}mg$

Where $(micro)_{k}$ is the coefficient of friction and m = 13.6 kg

$F_{k} = (0.16)(13.6)(9.8)\\$

$F_{k} = 21.32 \ N$

Now, Finding the net force:

$F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\$

$F_{net} = 28.7 \ N$

Finding Acceleration:

$a = \frac{F_{net}}{m}$

$a = \frac{28.7}{13.6}$

$a = 2.1 \ m/s^2$

8 0

3 years ago

Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

8 0

3 years ago

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu

galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

$F=I \times B \times L \times sin(\theta)$