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3 years ago

during a crash, the acceleration was less than 30g. Calculate the force on a 70 kg person accelerating at this rate.

Physics

1 answer:

Igoryamba3 years ago

4 0

Answer:

less than 20580 N

Explanation:

According to the newton's second law of motion

Force = mass * acceleration

(assuming gravitational acceleration =9.8 m/s2 )

acceleration = 30*9.8 = 294 m/s2

acting Force = 70 * 294

= 20580 N

Since the acceleration was less than 30g , acting force should also be less than 20580 N

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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

Round to the hundredths place.

Ivanshal [37]

Answer:

21 protons

Explanation:

4 0

3 years ago

Jackson throws a football 30 meters at a speed of 15 m/s. How long was the football in the air before Laurence caught it for tou

ch4aika [34]

Answer:

Explanation:

Given parameters:

Distance = 30m

Speed = 15m/s

Unknown:

Time before Laurence caught it = ?

Solution:

To solve this problem;

Speed = $\frac{disance }{time}$

Time taken = $\frac{distance }{speed }$ = $\frac{30}{15}$ = 2s

The time it takes is 2s

6 0

3 years ago

Read 2 more answers

An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla

dusya [7]

Answer:

C). $U_f = \frac{U_0}{2}$

Explanation:

As we know that capacitance of a given capacitor is

$C = \frac{\epsilon_0 A}{d}$

now we know that energy stored in the capacitor plates

$U_0 = \frac{Q^2}{2C}$

here if all the dimensions of the capacitor plate is doubled

then in that case

$C' = \frac{\epsilon_0 (4A)}{2d}$

here area becomes 4 times on doubling the radius and the distance between the plates also doubles

So new capacitance is now

$C' = 2C$

so capacitance is doubled

now the final energy stored between the plates of capacitor is given as

$U_f = \frac{Q^2}{2C'}$

so the final energy is

$U_f = \frac{Q^2}{4C}$

$U_f = \frac{U_0}{2}$

4 0

3 years ago

The diagram shows a position-time graph What is the displacement of the object

algol13

The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.

5 0

3 years ago