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inn [45]
3 years ago
5

Which gravitational force field dagram is drawn correctly?

Physics
2 answers:
VladimirAG [237]3 years ago
8 0
The correct diagram is B:)
Shalnov [3]3 years ago
7 0
The answer is B because no matter where you are on earth, gravity is pushing straight down towards the earth, so B
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I need help pleaseee
Ludmilka [50]

Answer:

8) 709.8875 J

9) The object is at 7.24375 m from the ground

10) Kinetic energy increases as the object falls.

Explanation:

We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:

h(t) = hi - (g/2) t^2

hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):

h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m

This is the position of the 10 kg object after 0.75 seconds (answer for part 9)

Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:

U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)

Part 10)

the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.

3 0
3 years ago
Boneless pizza boneless pizzaboneless pizza boneless pizzaboneless pizza boneless pizza
Musya8 [376]

Answer:

pizza dont got no bone in it

Explanation:

7 0
3 years ago
Find the height or length of these natural wonders in km, m, and cm.A. A cave system with a mapped length of 354 miles.B. A wate
cricket20 [7]

Explanation:

Some standard unit conversion are 1 mile = 1.609344 km, 1 ft.= 30.48 cm,

1 km= 1000 m or 1 m = 0.001 km and 1 m= 100 cm or 1 cm=0.01 m.

Now, use these values to convert the given lengths.

A. length of cave = 5 miles (given)

From standard value 1 mile = 1.609344 km

\Rightarrow 5 miles = \times 1.609344 km= 8.04672 km,

\Rightarrow 5 miles =8.04672 \times 1000 m= 8046.72 m [ as 1 km = 1000 m]

\Rightarrow 5 miles = 8046.72 \times 100 cm=804672 cm

B. Height of the waterfall = 1235 ft. (given)

1 ft.= 30.48 cm [ from standard value]

\Rightarrow 1235.2 ft.= 1235.2\times 30.48 cm=37648.896 cm,

\Rightarrow 1235.2 ft.=37648.896 \times 0.01 m =376.48896 m [ as 1 cm = 0.01 m]

\Rightarrow 1235.2 ft.=376.48896 \times 0.001 km =0.37648896 km [ as 1 m = 0.001 km]

C. Height of the mountain= 21320 ft. (given)

From standard value: 1 ft.= 30.48 cm

\Rightarrow 21320 ft.= 21320 \times 30.48 cm=649833.6 cm,

\Rightarrow 21320 ft.=649833.6 \times 0.01 m =6498.336 m [ as 1 cm = 0.01 m]

\Rightarrow 21320 ft.=6498.336 \times 0.001 km =6.498336 km [ as 1 m = 0.001 km].

D. Depth of canyon =6630 ft.

From standard value: 1 ft.= 30.48 cm

\Rightarrow 6630 ft.= 6630 \times 30.48 cm=202082.4 cm,

\Rightarrow 6630 ft.=202082.4  \times 0.01 m =2020.824  m [ as 1 cm = 0.01 m]

\Rightarrow 6630 ft.=2020.824  \times 0.001 km =2.020824  km [ as 1 m = 0.001 km].

4 0
4 years ago
A speeding car traveling at 24.8 m/s passes a police car that is at rest. The police car begins its pursuit at the instant the s
marusya05 [52]

Answer:

the acceleration required is 1.37m/s^2

Explanation:

The car is having a constant velocity movement, so if we calculate the time to reach 897m, we can use it to find the acceleration the policeman need to apply to reach the car.

x=v*t\\t=\frac{x}{v}\\t=\frac{897m}{24.8m/s}\\t=36.17s

the policeman is traveling with a constant acceleration starting from rest so:

x=\frac{1}{2}*a*t^2\\\\a=\frac{2*x}{t^2}\\\\a=1.37 m/s2

7 0
3 years ago
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Alekssandra [29.7K]

Answer:

time at which Kathy overtakes Stan is 6.70 sec

Explanation:

given data

time = 1 sec

acceleration= 3.4 m/s²

acceleration = 4.49  m/s²

to find out

the time at which Kathy overtakes Stan

solution

we consider here travel time for kathy = t1

and travel time for stan is = t2

and we know initial velocity = 0

so

t1 = 1 + t2

and distance travel equation by kinematic is

d1 = ut + \frac{1}{2} at²

d1 = 0+  \frac{1}{2} 4.49 (t2)²   ................1

and

d2 = 0 + \frac{1}{2} 3.4 (1+t2)²   ..................2

and when overtake distance same so  from equation 1 and 2

\frac{1}{2} 4.49 (t2)²  = \frac{1}{2} 3.4 (1+t2)²

t2 = 6.703828 sec

so time at which Kathy overtakes Stan is 6.70 sec

3 0
3 years ago
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