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Leto [7]
3 years ago
6

A dog chase is a cat until the dog gets tired. A graph of its velocity overtime is shown below. What is the dogs average acceler

ation from 0-12 seconds?

Physics
2 answers:
dsp733 years ago
8 0

Answer: 0.67 m/s^2

Explanation:

Bc it is trust me

Debora [2.8K]3 years ago
5 0

Answer:

.67 m/s^2

Explanation:

khan academy

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Why is the combination of two protons and two neutrons stable, but two protons and one neutron is not?
RideAnS [48]
Because it's unbalanced.
4 0
3 years ago
A police radar gun uses X-band microwave radiation at a frequency of 12.2 GHz. Microwaves travel at the speed of light, or 3x108
quester [9]

Answer:

A police radar gun uses X-band microwave radiation at a frequency of 13.1 GHz. Microwaves travel at the speed of light, or 3x108 m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car.

a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

ANSWER: 2533 Hz

Explanation:

6 0
3 years ago
A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total
Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

v_2 = v - 1

now we know that total time of the motion is 19 hr

so we will have

t_1 = \frac{90}{v}

t_2 = \frac{90}{v - 1}

so we will have

t_1 + t_2 = 19 hr

\frac{90}{v} + \frac{90}{v-1} = 19

90(2v - 1) = 19(v^2 - v)

19 v^2 - 199 v + 90 = 0

by solving above equation we have

v = 10 mph

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0
3 years ago
What is the final velocity of a body if it is moving with 13 m/s in 300 seconds and its acceleration is 30 m/s 2.
Nutka1998 [239]

Answer:

9013m/s

Explanation:

acceleration= v- u/ t

=> at = v-u

=> v = at + u

=> v =30*300+13

= 9013m/s

6 0
2 years ago
The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The
Vika [28.1K]

Answer:

No, there won't be a collision.

Explanation:

We will use the constant acceleration formulas to calculate,

v = u + a*t

0 = 25 + (-0.1)*t

t = 250 seconds (the time taken for the passenger train to stop)

v^2 = u^2 + 2*a*s

0 = (25)^2  + 2*(-0.1)*s

s = 3125 m (distance traveled by passenger train to stop)

If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur

Speed*time = distance

Distance = (15)*(250)

Distance = 3750 m

As the distance is way more, there won’t be a collision

5 0
3 years ago
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