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Alona [7]
3 years ago
13

If the brakes are applied and the speed of the car is reduced to 12 m/s in 15 s , determine the constant deceleration of the car

.
Physics
1 answer:
zavuch27 [327]3 years ago
5 0

Given:

\triangle v = 12 m/s

Time = 15 s

To find:

Deceleration = ?

Formula used:

Deceleration is given by,

a = \frac{\triangle v}{t}

Solution:

Acceleration of the body is rate of increase of velocity. Deceleration of the body is negative acceleration i. e rate of decrease of velocity.

Thus, deceleration of the body is given by,

a = \frac{\triangle v}{t}

where a is the deceleration of the body

v is the velocity

\frac{\triangle v}{t} is rate of change of velocity

t is the time

a = \frac{12}{15}

a = 0.8 m/ s^{2}

Thus, deceleration is 0.8 m/ s^{2}.

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football hemets have pads that are filled with air and thick foam so when they are hit the foam asorbs the hit and the air keeps the hard outer shell of the helmet from hiting the players head

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3 years ago
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An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1
hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity =  \frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}

Average velocity = < 0, 316000, 146000> m/s

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d =  < 0, 316000, 146000>  x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

5 0
4 years ago
A. Obtain the following: microwave, ruler, something meltable (e.g. candy bar, marshmallows) B. A microwave works by setting up
Makovka662 [10]

Answer:

E = 124.7 N / C

Explanation:

Let's analyze the exercise: the microwave creates an electromagnetic wave of frequency F = 2.45 GHz, this wave is introduced into the microwave cavity and is reflected on the metal walls, which is why one or more standing waves are formed.

The electric field of the standing wave is

            I = E²

            E =√I

where I is the intensity of the radiation.

What is it

             I = P / A

where P is the effective emission power, almost all the power of the microwave and A is the area of ​​the cavity, in the most used microwaves

P = 700 W and the area is A = 25 x 18 cm² = 0.045 m²

             I = 700 / 0.045

             I = 15555.56 W/m²

let's calculate the electric field

            E = √15555.56

            E = 124.7 N / C

7 0
3 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

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