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dalvyx [7]
3 years ago
7

You are tossing pizza dough to get it ready to bake. If you toss the dough upward with a speed of 3.1 m/s, how high does it go?

Physics
2 answers:
ella [17]3 years ago
7 0
0.49
hope it help Apex 







belka [17]3 years ago
3 0

Answer:

It goes to the height of 0.50 m.

Explanation:

Given that,

Initial velocity = 3.1 m/s

We need to calculate the height

Firstly, We calculate the time

Using equation of motion

v=u+gt

t=\dfrac{v-u}{g}

Where, v = final velocity

u = initial velocity

t = time

g = acceleration due to gravity

Put the value in the equation

t=\dfrac{0-3.1}{-9.8}

t =0.32\ sec

Now, We calculate the height

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, s = height

t = time

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.32)^2

s=0.50\ m

Hence, It goes to the height of 0.50 m.

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Plz help ASAP I'll mark as brainliest ​
gogolik [260]

Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

8 0
3 years ago
. How does the intensity of sunlight received affect Earth’s temperatures?
leonid [27]
When the sun's rays strike Earth's surface near the equator, the incoming solar radiation is more direct (nearly perpendicular or closer to a 90˚ angle). Therefore, the solar radiation is concentrated over a smaller surface area, causing warmer temperatures.
7 0
3 years ago
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
SVETLANKA909090 [29]

Answer:

The  value is  v =  -0.04 \  m/s

Explanation:

From the question we are told that

   The  mass  of the block is  m  =  2.0 \ kg

   The  force constant  of the spring is  k  =  590 \ N/m

   The amplitude  is  A =  + 0.080

   The  time consider is  t =  0.10 \  s

Generally the angular velocity of this  block is mathematically represented as

      w =  \sqrt{\frac{k}{m} }

=>   w =  \sqrt{\frac{590}{2} }

=>   w = 17.18 \  rad/s

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         v  =  -A w sin (w* t )

=>       v  = -0.080 * 17.18 sin (17.18* 0.10 )

=>       v =  -0.04 \  m/s

7 0
4 years ago
Where are the sun's rays the strongest, which makes the ocean water the warmest ?
IgorC [24]
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8 0
3 years ago
A car can go from 0 m/s to 38 m/s in 4.5 seconds. If a net force of 6570 N acted on
masha68 [24]

The Mass of the car = 782.1 Kg

<h3>What is the mass of the car?</h3>

The mass of the car is calculated as follows:

  • Mass = Force/ acceleration

The force on the car = 6570 N

The acceleration of the car, a = 38 - 0/4.5

acceleration = 8.44 m/s²

Mass of the car = 6570/8.44

Mass of the car = 782.1 Kg

In conclusion, the mass of the car is obtained from the acceleration and force on the car.

Learn more about mass and acceleration at: brainly.com/question/19385703

#SPJ1

3 0
2 years ago
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