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Mamont248 [21]
3 years ago
7

What symbols are used to explain displacement in physics​

Physics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

In this text the upper case Greek letter Δ (delta) always means “change in” whatever quantity follows it; thus, Δx means change in position. Always solve for displacement by subtracting initial position x0 from final position xf

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A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions
erastova [34]
•wind
•snow
•high tide/low tide
•thunder/lightning storms
5 0
3 years ago
A girl is bouncing on a trampoline where is her gravitational potential energy a maximum and where is her kinetic energy maximum
Stella [2.4K]

Answer:

When you jump down, your kinetic is converted to potential energy of the stretched trampoline. The trampoline's potential energy is converted into kinetic energy, which is transferred to you, making you bounce up. At the top of your jump, all your kinetic energy has been converted into potential energy. Right before you hit the trampoline, all of your potential energy has  been converted back into kinetic energy. As you jump up and down your kinetic energy increases and decrease.

7 0
3 years ago
How much work is done against gravity when lowering a 16 kg box 0.50 m? (g = 9.8 m/s2)
leonid [27]

Answer:

The work done against gravity is 78.4 J

Explanation:

The work is calculated by multiplying the force by the distance that the

object moves

W = F × d, where W is the work , F is the force and d is the distance

The SI unit of work is the joule (J)

We need to find the work done against gravity when lowering a

16 kg box 0.50 m

→ F = mg

→ m = 16 kg, and g = 9.8 m/s²

Substitute these value in the rule

→ F = 16 × 9.8 = 156.8 N

→ W = F × d

→ F = 156.8 N and d = 0.50

Substitute these values in the rule

→ W = 78.4 J

<em>The work done against gravity is 78.4 J</em>

6 0
3 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

W=m(y)g\times dy

At any position 'y' the weight shall be sum of weft of water and weight of string

\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y

Thus applying values we get

W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J

8 0
3 years ago
Ma puteti ajuta sa rezolv la fizica o problema ?
xxTIMURxx [149]
Care este problema? Btw why are you speaking Romanian 
7 0
3 years ago
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