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Artist 52 [7]
3 years ago
8

Classify the waves as longitudinal or transverse.

Physics
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

1. transverse

2. transverse

3. longitudinal

4. idk

Explanation:

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The stopping distances associated with slower speeds approximate the forward visibilities provided by low beam lights. However,
aleksandr82 [10.1K]

Answer:

a. True

Explanation:

Illumination distance is the distance, up to which the light of the vehicle can reach. Hence, it is a maximum distance from the, that driver can see.

Stopping distance is the minimum distance required by the car to stop after brakes are applied.

So, in order to avoid any accident the illumination distance must be greater than the stopping distance. So, the driver can stop the vehicle in time, when he sees something in front of it.

Since, the stopping distance in this case is two or three times longer than illumination distance. Therefore, low beam light does not provide enough visibility in high speed driving situations.

Hence, the correct option is:

<u>a. True</u>

<u></u>

5 0
3 years ago
What is acceleration
kap26 [50]

Answer:

The rate of change of velocity is acceleration. It's SI unit is m/s².

6 0
2 years ago
Am arrow of mass 1000kg is shot into a wooden block of mass 5000lg lying at rest in a smooth surface.If the arrow travels 15m/s
Ber [7]

Answer:

Vf=3

Explanation:

you must first write your data

data before impact

M1=1000 M2=5000

V1=0 m/s V2 =0m/s

data after impact

M1=1000 M2=5000

V1=15m/s V2=?

M1V1 +M2V2=M1V1 +M2V2f

(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf

0=15000+5000Vf

- 15000÷5000=5000Vf÷5000

Vf= -3

Vf =3

6 0
2 years ago
What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?
sergey [27]

Answer:2.5m

Explanation:

3 0
1 year ago
An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
dsp73

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

3 0
2 years ago
Read 2 more answers
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