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Ulleksa [173]
3 years ago
11

9. A car driver brakes gently. Her car slows down front --

Physics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

b)

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

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A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t
Tom [10]

Answer: Yes, he is exceeding the speed limit

Explanation:

Hi!

This is problem about unit conversion

1 mile = 1,609.344 m

Then the speed limit v is:

v = 75 mi/h = 120,700.8 m/h

1 hour = 60 min = 60*60 s = 3,600 s

v = (120,700.8/3,600) m/s = 33.52 m/s

38 m/s is higher than the speed limit v.

5 0
3 years ago
Gumuhit ng paglalaruan ng<br>agawang sulok at ibigay ang bawat<br>sukat nito.​
Anarel [89]

Answer:

there's a link to answer

Explanation:

there's a link to answer in another's persons response it helps a lot just click on it

8 0
3 years ago
A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki
Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

\tau=I*\alpha

given

\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?

\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2

the angular velocity is

\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s

KE= 1/2*I* \omega^2

KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J

4 0
3 years ago
.
beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s)  = 9m

Time (t)  = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0
3 years ago
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.  

8 0
3 years ago
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