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3 years ago

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1. Use Newton's first law of motion to explain how wearing a seat belt in a moving car could help prevent injury

1 answer:

hjlf3 years ago

4 0

The diver is plummiting to earth and if he does not put it soon he is going to die

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PLEASE HELP Due Soon!

Taya2010 [7]

Answer:

Explanation:

When objects collide, energy can be transferred from one object to another, thereby changing their motion. In such collisions, some energy is typically also transferred to the surrounding air; as a result, the air gets heated and sound is produced.

8 0

3 years ago

Ultrasound involves high-frequency sound waves. Sound waves can be used to image internal structures. This is the basis of

Naddik [55]

Answer:

tomography

Explanation:

3 0

3 years ago

Read 2 more answers

A block of mass 3.1 kg, sliding on a horizontal plane, is released with a velocity of 2.3 m/s. The blocks slides and stops at a

jeka57 [31]

To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity

Our values are given as,

$\text{Initial speed} =V_i = 2.3 m/s$

$\text{Final speed}= V_f = 0 m/s$

$\text{Stopping distance = }d = 1.9 m$

$a = acceleration$

$\text{mass} = m = 3.1kg$

Using the kinematic equation of motion we have

$V_f^2 = V_i^2 + 2 a d$

$0^2 = 2.3^2 + 2 a (1.9)$

$a = -1.39211 m/s^2$

Now if the initial velocity is quadrupled we have that,

$\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s$

$\text{Final speed}= V_f' = 0 m/s$

$\text{Stopping distance = }d'$

$V_f^2' = V_i^2' + 2 a d'$

Replacing the values

$0^2 = 9.2^2+ 2 (-1.39211) d'$

$d' = 30.44m$

Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.

3 0

3 years ago

Please tell me the answer

maria [59]

Answer:

for which one

Explanation:

4 0

3 years ago

In an RC series circuit, ℰ = 17.0 V, R = 1.60 M, and C = 1.80 µF.

LiRa [457]

Answer:

(a) τ=2.88 s, (b) q₀= 30.6μC and (c) t=1.434s

Explanation:

A RC circuit is an resistor(R)-capacitor(C) electric circuit.

(a) In a resistor-capacitor circuit, the time constant (τ) can be calculated by:

$\tau = RC$

<em>where R: is the resistence and C: the capacitance of the capacitor</em>

$\tau = (1.60\cdot 10^{6} \cdot 1.80\cdot 10^{-6} = 2.88 s$

(b) The maximum charge (q₀) is giving by:

$q_{0} = \epsilon \cdot C$

<em>where ε: is the voltage across the capacitor</em>

$q_{0} = 17.0 V \cdot 1.80 \cdot 10^{-6} F = 3.06 \cdot 10^{-5} C = 30.6 \mu C$

(c) The time (t) that take the charge (q) to build up to 12 μC can be calculated from the next equation:

$q = q_{0}(1 - e(\frac{-t}{RC}))$

$t = RC Ln( \frac{q_0}{q_0 - q}) = \tau Ln( \frac{q_0}{q_0 - q})$

$t = 2.88 \cdot Ln (\frac{30.6 \cdot 10^{-6}}{(30.6 \cdot 10^{-6} - 12.0 \cdot 10^{-6}})$

$t = 1. 434 s$

Have a nice day!

5 0

3 years ago