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BabaBlast [244]

3 years ago

9

during a test crash an airbag inflates to stop a dummies forward motion. the dummies mass is 75 kg. if the net force on the dumm

y is 825N toward the rear of the car what is the dummies deceleration

2 answers:

marta [7]3 years ago

6 0

Answer:

Dummies deceleration is 11m/s²

Explanation:

According to Newton's second law, the relationship between the force applied to an object, its mass, and the acceleration (or deceleration) caused, is given by:

$F=ma$

Solving for acceleration:

$a=\frac{F}{m}$

If we plug in the vales (taking into account the negative sign of the force, as its directed backwards):

$a=\frac{-825N}{75kg}=-11m/s^2$

$a=-11m/s^2$

Deceleration is simply the magnitude of negative acceleration. Therefore, dummies deceleration is 11m/s²

den301095 [7]3 years ago

5 0

Use Force=Mass x Acceleration (newtons second law states force is directly proportional to the acceleration) so you can say that the force is negative and solve for Acceleration.

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Answer:

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2 years ago

If there are two photons with different energies, the one that has a higher energy (a) has the higher frequency (b) has the high

kozerog [31]

Answer:

(a) has the highest frequency

Explanation:

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4 0

3 years ago

an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object

emmainna [20.7K]

Answer:

The acceleration of the object is $-30\ m/s^2$

Explanation:

Given:

Initial velocity of object $v_i$ = 200 feet/second

Final velocity of object $v_f$ = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration $a$ is given by:

$a=\frac{v_f-v_i}{t}$

where $v_f$ represents final velocity, $v_i$ represents initial velocity and $t$ is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a=-30\ m/s^2$

The acceleration of the object is $-30\ m/s^2$ (Answer). The negative sign shows the object is slowing down.

4 0

4 years ago

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

3 years ago