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KengaRu [80]
3 years ago
12

Could I get help on this question please . My parents won’t help me /:

Physics
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

From the question given above, the following data were obtained:

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

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In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? pls help
Alexxx [7]

Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

4 0
3 years ago
Read 2 more answers
A skateboarder rolls down the sidewalk with an initial velocity of 0 m/s. If her acceleration is
VikaD [51]

Answer:

80m/s

Explanation:

to find it you have to work it out by using the formula distance divided by speed to find time.

3 0
2 years ago
A deciliter is how many times larger than a millimeter?
jarptica [38.1K]
Milliliter  centiliter  deciliter  liter   dekaliter   hectoliter kiloliter

All related by 10's   every move to the right is 10x larger than the one to its left

Since deciliter is two steps away from milliliter it is 10 x 10 or 100 times as large.

If the question is meant to be a trick then the answer is 99x larger.  

It should read "A deciliter is how many milliliters"  ofr "a milliliter is how much of a deciliter?"
 Once you say larger than you could be confusing subtraction with multiplication.  How much larger than 30 is 3?  Answer is clearly 27.  How many times as large as 3 is 30?  Answer is clearly 20. How many times larger than 3 is 30?  Hmmmm?  Which one of the two does he mean?
I am sure your teacher meant you to consider multiplication, but just in case, I included the "trick" answer.
8 0
3 years ago
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is
natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

C=\frac{\epsilon _{0}A}{d}  

Here, C is the capacitance, \epsilon _{0} is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

C=\frac{K\epsilon _{0}A}{d}

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

E=\frac{CV^{2}}{2}

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0
2 years ago
Help please. Will give brainliest
solmaris [256]

Answer:

Your answer would be:
B.) Both carry energy in the form of vibration



Explanation:

Have a great rest of your day
#TheWizzer

4 0
2 years ago
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