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Citrus2011 [14]
3 years ago
9

Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope

at a constant speed of 10 m/s. What is the net force on the cart during this motion
Physics
1 answer:
VladimirAG [237]3 years ago
6 0

Car is moving on the glassy slope with constant speed

Now we know that

a = \frac{dv}{dt}

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

a = 0

now as we know by Newton's II law

F = ma

since a = 0

F = 0

so net force will be ZERO on it during this motion

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Order the speed of sound through these materials from the slowest to the fastest.
Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0
3 years ago
There are 8.8 liters of gas in a piston at a pressure of 1.75 atmospheres. The temperature remains constant, and the gas is comp
saveliy_v [14]

The new pressure P2 is 2.48 atmosphere.

<u>Explanation:</u>

Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.

By using Boyles's law,

               pressure is inversely proportional to volume,

                                         P1 V1 = P2 V2

where P1, V1 represents the first pressure and volume,

           P2, V2 represents the second pressure and volume

                                              P2 = (P1 V1) / V2

                                                    = (1.75 \times 8.8) / 6.2

                                              P2  = 2.48 atmosphere.  

         

5 0
3 years ago
What is the momentum of a 0.15 kilgram baseball moving at 20 m/s?
natita [175]
Momentum = mv
= .15 *20
= 3 kgm/s^2
5 0
3 years ago
A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next,
Tamiku [17]

Answer:

Explanation:

Let 100 m/s  be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

Vertical component = - 100 sin 42 . in upward direction

66.91 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = - 66.91 t  + .5 x 9.8 x t²

4.9 t² -  66.91 t - 2.1 = 0

t = 13.685 s

Horizontal distance covered

= 13.685 x 74.31

= 1016.93 m

If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

Vertical component = - 100 sin 42 . in upward direction

64.27 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = -76.60 t  + .5 x 9.8 x t²

4.9 t² -  76.60 t - 2.1 = 0

t = 15.659  s

Horizontal distance covered

= 15.659 x 76.60

= 1199.49  m

So horizontal range is increased , if angle of projection is increased .

8 0
3 years ago
WILLL GIVE 5 STARS BRAINIEST AND THANKS AND 20 POINTS EACH ANSWER!!!!
vekshin1

Answer:

The time she takes to reach the water from when she jumps off the platform is 1.71 s

Explanation:

According to the equations of motion, we have;

v = u - g·t

v² = u² - 2·g·s

s₁ = u₁·t₁ + 1/2·g₁·t₁²

The given parameters are;

The height of the platform (assumption: above the water) = 10 m

The velocity with which she jumps, u = 3 m/s

The acceleration due to gravity, g = 9.81 m/s²

The height to which she jumps, s, is found as follows;

v² = u² - 2·g·s

At maximum height, v = 0, which gives;

0 = 3² - 2×9.81×s

2×9.81×s = 3² = 9

s = 9/(2×9.81) = 0.4587 m

s = 0.4587 m

The time to maximum height, t, is found as follows;

v = u - g·t

0 = 3 - 9.81×t

9.81×t = 3

t = 3/9.81 = 0.3058 s

The total distance, s₁ from maximum height to the water surface = s + 10 = 0.4587 + 10 = 10.4587 m = 10.46 m

The time to reach the water from maximum height, t₁, is found as follows;

s₁ = u₁·t₁ + 1/2·g₁·t₁²

Where;

s₁ =  The total distance from maximum height to the water surface = 10.46 m

u₁ = The initial velocity, this time from the maximum height = 0 m/s

g₁ = The acceleration due to gravity, g (positive this time as the diver is accelerating down) = 9.81 m/s²

t₁ = The time to reach the water from maximum height

Substituting the values gives;

s₁ = u₁·t₁ + 1/2·g₁·t₁²

10.46 = 0·t₁ + 1/2·9.81·t₁²

t₁²= 10.46/(1/2×9.81) = 2.13 s²

t₁ = √2.13  = 1.46 s

Total time = t₁ + t = 1.46 + 0.3058 = 1.7066 ≈ 1.71 s.

Therefore, the time she takes to reach the water from when she jumps off the platform = 1.71 s.

8 0
3 years ago
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