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2 years ago

PLEASE HELP!

Physics

1 answer:

Keith_Richards [23]2 years ago

8 0

Wavelength if it’s wrong pls don’t hate me
And if you think I should get branliest feel free to give it to me

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Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov

Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

$K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}$

Final kinetic energy is :

$K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}$

Now, fraction of initial kinetic energy loss is :

$Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25$

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0

2 years ago

The distance from earth to teh sun usb approxximately 93 million miles. a scientist would write that number as

erastova [34]

A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as 1.49 x 10⁸ kilometres .)

6 0

3 years ago

A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What

tangare [24]

Answer:

man will move in opposite direction with speed

$v_1 = 1.66 \times 10^{-3} m/s$

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

$P_i = P_f$

$0 = m_1v_1 + m_2v_2$

here we have

$0 = (97)v_1 + 0.062(2.6)$

$v_1 = -\frac{0.1612}{97}$

$v_1 = -1.66 \times 10^{-3} m/s$

3 0

2 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago

What Are Three Characteristic Properties Of Ionic Compounds

sashaice [31]

Ionic bonds have...

- High melting points
- Conductive properties
- Are good insulators
(Extra: Form crystal-like structures over just plain molecules!)

7 0

3 years ago