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3 years ago

11

6. Rock Ais thrown horizontally off a cliff with a velocity of 15 m/s. The rock lands 75m from

1 answer:

Harman [31]3 years ago

3 0

Answer:

5 seconds

Explanation:

You might be interested in

A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?

pochemuha

10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv

7 0

4 years ago

Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require

strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

Em₀ = U = m g h

Final point. In the highest part of the loop

Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

I = m R²

v = w R

we substitute

Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

em_f = m v² + 2 m g R

Energy is conserved

Emo = Em_f

mgh = m v² + 2m g R

h = v² / g + 2R

The lowest velocity that the ball can have at the top of the loop is v> 0

h> 2R

3 0

3 years ago

The diagrams show objects' gravitational pull toward each

zaharov [31]

I don’t know sorry then ya soo

4 0

3 years ago

Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%

djyliett [7]

Answer:

$6.8370869499\times 10^{20}\ N$

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

e = Charge of electron = $1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

$q=imbalance\times n\times e$

Total number of protons and electrons in each sphere

$n=\dfrac{mN_AZe}{M}$

$q=imbalance\times \dfrac{mN_AZe}{M}$

$q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C$

$q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C$

Electrical force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N$

The electrostatic force of attraction between them is $6.8370869499\times 10^{20}\ N$

4 0

4 years ago

Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

3 years ago