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Natalija [7]
3 years ago
9

During a race, a runner runs at a speed of 6 m/s. Four seconds later, she is running at a speed of 10 m/s. what is the runners a

cceleration? show your work.
Physics
1 answer:
Georgia [21]3 years ago
8 0
<span>a = (v2 - v1)/t = (10 - 6)/2 = 2 m/sec/sec (average acceleration)</span>
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How many joules of work are done against a truck when a force of 50 N pushes it 1 kilometer away
ELEN [110]

Answer:

Work = 50,000 J

Explanation:

Work = force * distance

Given that,

  • force = 50N
  • Distance = 1km = 1000m

Work = 50 * 1000

Work = 50,000 J

8 0
2 years ago
Which of the following would produce the most power?
Fantom [35]

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

Power = \frac {10*9.8*10}{5} = 490 Watts

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

6 0
2 years ago
The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small c
Elodia [21]

Answer:

the distance between adjacent fringes is increased by a factor o 2

Explanation:

To find how the distance between fringes is modified you can use the following formula for the calculation of the distance between fringes:

\Delta y=\frac{\lambda D}{d}

D: distance to the screen

d: distance between slits

λ: wavelength of the light

if d is decreased by a factor of 2, that is d'=1/2d, you have:

\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y

hence, the distance between adjacent fringes is increased by a factor o 2

4 0
3 years ago
The velocity of a car changes from 20 m/s east to 5 m/s east in 5 seconds. What is the acceleration of the car?
Klio2033 [76]
acceleration =  \frac{ v_{2}- v_{1}  }{t} = \frac{5-20}{5} =-3m/s^{2}
3 0
3 years ago
A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 3.00 s.
Misha Larkins [42]

Answer:

Explanation:

Initial velocity of mailbag u = 2 m/s

acceleration downwards a = g = 9.8 m/s²

time t = 3 s

a ) final velocity v = ?

v = u + at

= 2 + 9.8 x 3

= 31.4 m /s

b )

s = ut + 1/2 g t²

s is relative displacement of mailbag

u = relative initial velocity of mailbag = 0

relative acceleration = g = 9.8 m /s²

time t = 3 s

s = 0 + 1/2 x 9.8 x 3²

= 44.1 m

relative displacement of mailbag = 44.1 m .

4 0
2 years ago
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