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denis-greek [22]
2 years ago
14

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. even without knowing how bi

g the pressure cooker is, or what altitude it is being used at, we can make predictions about how much force the lid will experience under different conditions.
Physics
1 answer:
ankoles [38]2 years ago
5 0
Even without knowing the size of the pressure cooker or the altitude at which the pressure cooker will be used, we can easily determine the force on the lid by experience because just as boiling water, the amount of pressure inside the vessel (in this case, the cooker) increases as the temperature increases. 
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A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo
NISA [10]

Explanation:

The given data is as follows.

    Mass of small bucket (m) = 4 kg

    Mass of big bucket (M) = 12 kg

    Initial velocity (v_{o}) = 0 m/s

    Final velocity (v_{f}) = ?

  Height H_{o} = h_{f} = 2 m

and,    H_{f} = h_{o} = 0 m

Now, according to the law of conservation of energy

         starting conditions = final conditions

  \frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}

     \frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)

                 235.44 = 8V^{2}_{f} + 78.48

                V_{f} = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0
1 year ago
A spaceship is travelling at 20,000.0 m/s. After 5.0 seconds, the rocket thrusters are turned on. At the 55.0 second mark, the s
tankabanditka [31]

Answer:

80 m/s^2

Explanation:

The acceleration of an object is given by:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the rocket in this problem,

u = 20,000 m/s

v = 24,000 m/s

t = 55.0 - 5.0 = 50.0 s

Substituting,

a=\frac{24000-20000}{50}=80 m/s^2

7 0
2 years ago
Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th
katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0
2 years ago
Read 2 more answers
Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a
mote1985 [20]

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0
1 year ago
A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor
nalin [4]

Answer:

u_x=38.13\ m/s

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

S=u_yt+\dfrac{1}{2}gt^2

Here u_y=0

S= 57 cm

0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2

t=0.34 s

Now in the horizontal direction

x=u_xt

Here x=13 m

t= 0.34 s

So

13=u_x\times 0.34

u_x=38.13\ m/s

So the initial speed of ball is 38.13 m/s.

7 0
1 year ago
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