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3 years ago

7. Which law describes when a person lands on a

Physics

2 answers:

marissa [1.9K]3 years ago

6 0

Answer:

Newton's Third Law

Explanation:

Newton's third law

Newton's third law: “for every action, there is an equal and opposite reaction.” This is where you get the bounce. When you push down on the trampoline (or fall downward onto the trampoline bed), Newton's third law says that an equal and opposite reaction pushes back.

Anna007 [38]3 years ago

5 0

Answer:

It is Newton's 3rd law

Explanation:

Because: For every action force: (person jumping on a trampoline) there is a equal yet opposite reaction. Reaction force: (The trampoline pushed up against the person's feet.)

Hope this helped!

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Bob is moving at 0.967c with respect to Alice. At the exact instant he passes Alice, she fires a very short laser pulse in the s

yulyashka [42]

Explanation:

Speed of Bob, v = 0.967 c

At the exact instant he passes Alice, she fires a very short laser pulse in the same direction Bob is moving.

(a) We need to find the distance measured by Alice between Bob and the laser pulse. It is given by :

$d=ct-vt$

$d=t(c-v)$

$d=5.59\times (c-0.967c)$

$d=5.53\times 10^7\ meters$

(b) Distance measured by Bob between himself and the laser pulse is given by :

$d_B=ct$

$d_B=3\times 10^8\times 5.59$

$d_B=6.67\times 10^9\ meters$

Hence, this is the required solution.

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3 years ago

A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made

olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

$B=\dfrac{\mu_oI}{2\pi r}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}$

B = 0.000002 T

$B=10^{-5}\ T$

(2) Exact formula:

$B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }$

$B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }$

B = 0.00000199 T

B = 0.000002 T

Hence, this is the required solution.

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3 years ago

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of t

Firlakuza [10]

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

Voltage, V = 6 V, Resistance, R = 550 Ω

Current, I $=\frac{6}{550}=0.011A=11mA$

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) $\texttt{Power=}\frac{Energy}{Time}=\frac{0.066}{0.65\times 10^{-3}}=101.54W$

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3 years ago

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Maru [420]

The correct answer is D

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3 years ago

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A 930-kg sports car collides into the rear end of a 2000-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

bekas [8.4K]

Answer:2.103 m/s

Explanation:

Given

mass of sports car $m=930\ kg$

mass of SUV $M=2000\ kg$

Suppose u is the velocity if sports car before collision

Conserving momentum we get

$mu=(M+m)v$

$v=\dfrac{2000+930}{930}\times u$

$v=3.15\cdot u$

After collision the combined mass drag 2.8 m and finally stops

From work energy theorem work done by friction is equal to change in kinetic energy of the combined mass system

$\frac{1}{2}(M+m)v^2=\mu (M+m)gx$

where $\mu =\text{coefficient of friction}$

$x=\text{drag distance}$

$v=\sqrt{2\mu gx}$

$v=\sqrt{2\times 0.8\times 9.8\times 2.8}$

$v=6.626\ m/s$

Initial velocity $u=\frac{6.626}{3.15}$

$u=2.103\ m/s$

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3 years ago