A rock is dropped from the top of a building. after 4 seconds it is moving at 40 m/s. what was the acceleration of the rock

ivann1987 [24]

<span>
be aware of large trucks on the roadway
Tip: usually, these signs are hazard signs or warning.
Also, this isnt physics </span><span />

5 0

2 years ago

Read 2 more answers

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

2 years ago

If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0

2 years ago

Question 1 of 7What is the change in internal energy if 90 J of thermal energy is added to asystem, and the system does 30 J of

Dmitriy789 [7]

Given:

The thermal energy added to the system is Q = 90 J

The work done by the system on the surroundings is W = 30 J

To find the change in internal energy.

Explanation:

According to the first law of thermodynamics, the change in internal energy can be calculated by the formula

$\Delta U=\text{ Q-W}$

On substituting the values, the change in internal energy will be

$\begin{gathered} \Delta U\text{ =90-30} \\ =60\text{ J} \end{gathered}$

Final Answer: The chage in internal energy is 60 J (option D)

5 0

6 months ago

A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. If the charge on each plate has a magni

barxatty [35]

Answer:

The potential difference across the plates is 226 V.

Explanation:

Given;

area of the capacitor plate, A = 0.2 m²

separation, d = 0.1 mm = 0.1 x 10⁻³ m

charge on each plate, Q = 4 x 10⁻⁶ C

Charge on the capacitor is given by;

Q = CV

Where;

C is the capacitance of the capacitor, given as;

C = ε₀A / d

Then, the potential difference across the plates is given by;

$V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V$

Therefore, the potential difference across the plates is 226 V.

5 0

2 years ago