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A hungry hawk was preying on a lizard who was running northwards to get away from the low-flying hawk. If the lizard can run 8m

Anastaziya [24]

Answer:

2m/s²

Explanation:

velocity = displacement (distance in a specified direction /time

8 0

3 years ago

A car is running at a velocity of 50 miles/hour and the driver accelerates the car by 10miles/hour.How far the car travels from

dezoksy [38]

Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
Time t = 2 hours
Distance travelled S = ut + (at^2)/2
Substituting the values in the second equation of motion,
S = 50*2 + (10 * 2 *2)/2
S = 100 + 20
S = 120 miles
Therefore the distance travelled by the car in the next two hours is 120 miles

4 0

3 years ago

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A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a

Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0

3 years ago

1The density of an of an object with a volume of 60.0 and mass of 400.0g is______

Kobotan [32]

D is the answer
To get the density you divide mass by volume
So the equation is 400/60=d

4 0

3 years ago

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

Ivan

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy

Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

$h\nu = W + KE$

here W = work function of metal which shows the energy required to eject out electrons from metal surface

KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same

8 0

3 years ago