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Finger [1]
3 years ago
10

A baseball weighing 0.5 kg falls from the sky. You hit it with a bat with a force of 50 N at an upward angle of 45°.

Physics
1 answer:
kirill [66]3 years ago
6 0

Answer:

bchfdfu

<em>hjoufu</em><em> </em><em>I'm</em><em> not</em><em> sure</em><em> if</em><em> you</em><em> are</em><em> still</em><em> interested</em><em> in</em><em> the</em><em> position</em><em> and</em><em> would</em><em> like</em><em> to</em><em> know</em><em> if</em><em> you</em><em> are</em><em> interested</em>

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Explain other learning or ideas you gained about conservation of energy
Sav [38]

Answer:

It can help individuals understand concepts such as the water-energy nexus and how water and energy work together to create a higher quality of life. Smart Energy Education teaches students about renewable energy, energy conservation, and the importance of energy as we transition into the future.

<em>8 Easy Ways for Kids to Conserve Energy at Home and School</em>

Take a Shower Instead of a Bath. ...

Limit Electronic Usage. ...

Turn Everything off When You Aren't in the Room. ...

Keep Windows and Blinds Closed. ...

Plant a Tree. ...

Ride a Bike or Walk to School. ...

Use Both Sides of a Piece of Paper. ...

Use a Reusable Water Bottle.

4 0
2 years ago
Ship A is located 4.2 km north and 2.7 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a vel
kotykmax [81]

Answer:

(a) The x-component of velocity is 31.55 km/h

(b) The y-component of velocity is 44.92 km/hr

Solution:

As per the solution:

The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.

Velocity of ship A, \vec{u_{A}} = 22 km/h towards South = - 22\hat{j}

Velocity of ship B, \vec{u_{B}} = 39 km/h Towards North east at an angle of 36^{\circ} = \vec{u_{B}} = 39sin36^{\circ} \hat{j}

Now, the velocity of ship A relative to ship B:

\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}

\vec{u_{A}} = - 22\hat{j}

\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

Now,

\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}

\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}

4 0
3 years ago
A wind powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of win
Hunter-Best [27]
I'd go for b) v squared.
Wind is air in motion. So, it has kinetic energy. For a given mass of air at a certain speed, the kinetic energy would be (1/2)mv^2.
Since everything else in the chain seems to be proportional, then it's the v^2 bit that seems to be important. Hence the answer I post here.
7 0
4 years ago
What is the difference between engineering controls and administrative controls?
nadezda [96]

Engineering controls could involve altering the weight of the products, adjusting the heights of the work surfaces, or getting lifting equipment. Administrative controls are workplace policies, practices, and processes that reduce the likelihood of risky situations for employees.

<h3>What are Engineering controls?</h3>
  • Engineering controls are methods for preventing dangerous situations for workers by erecting a barrier between them and the danger or by ventilating the area to remove any hazardous materials.
  • Instead of depending on employee behavior or mandating that employees wear safety gear, engineering controls entail making physical changes to the workplace itself.
<h3>What is Administrative controls?</h3>
  • Administrative controls are policies, procedures, shift designs, or training that reduce the threat of a hazard to a person.
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Learn more about Administrative controls here:

brainly.com/question/17628591

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5 0
2 years ago
A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

<u>F = 51.3°</u>

8 0
3 years ago
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