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dybincka [34]
3 years ago
14

Plz plz help ...i really need steps for this answer

Physics
2 answers:
djyliett [7]3 years ago
8 0

Answer:

3.9 seconds

Explanation:

A man standing near a well throws a stone upwards with a velocity of 15ms^-1. The time the stone takes to reach the bottom of the well, which is 15m deep is  3.9 seconds.

y = y₀ + v₀ t + 1/2 at²

bija089 [108]3 years ago
6 0

Answer:

3.9 seconds

Explanation:

Use constant acceleration equation:

y = y₀ + v₀ t + ½ at²

where y is the final position,

y₀ is the initial position,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

Given:

y = 0 m

y₀ = 15 m

v₀ = 15 m/s

a = -9.8 m/s²

Substituting values:

0 = 15 + 15t + ½ (-9.8) t²

0 = 15 + 15t − 4.9t²

0 = 4.9t² − 15t − 15

Solve with quadratic formula:

t = [ -b ± √(b² − 4ac) ] / 2a

t = [ 15 ± √((-15)² − 4(4.9)(-15)) ] / 2(4.9)

t = [ 15 ± √(225 + 294) ] / 9.8

t = (15 ± √519) / 9.8

t = -0.79 or 3.9

It takes 3.9 seconds for the stone to reach the bottom of the well.

The negative answer is the time it takes the stone to travel from the bottom of the well up to the top of the well.

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Two heavy objects of masses 1 kg and 2 kg are moving towards each other under
SOVA2 [1]

The acceleration of the body in terms of the gravitational constant G is G.

According to Newton's law of universal gravitation;

F = Gm1m2/r^2

G = gravitational constant

m1 = mass of the first body

m2 = mass of the second body

r = distance between the two bodies

Substituting values to find the force on the two bodies;

F = G × 1  × 2/1^2

F = 2G

For the 2 Kg mass

F = ma

m = mass

a = acceleration

F = gravitational force

Hence,

2G = 2a

a = 2G/2

a = G

Learn more: brainly.com/question/13860566

5 0
3 years ago
Please help with Physics Circuits!
Zigmanuir [339]
1) Let's start by calculating the equivalent resistance of the three resistors in parallel, R_2, R_3, R_4:
\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}
From which we find
R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega
So, the correct answer is D) 


2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega

And these are connected in series with a resistor of 10 \Omega, so the equivalent resistance of the circuit is
R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega

And by using Ohm's law we find the current in the circuit:
I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A
So, the correct answer is C).


3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega

And by using Ohm's law we find the current flowing in the circuit:
I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V
So, the correct answer is D).


4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}
From which we find
R_{23} = 7.55 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is:
R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega

The current in the circuit is given by Ohm's law
I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A

Now we can compare the voltage drops across the resistors. Resistor 1:
V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V
Resistor 4:
V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V

So, the greatest voltage drop is on resistor 1, so the correct answer is D).


5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
P_3 = I_3^2 R_3 =  \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.
5 0
3 years ago
Read 2 more answers
Question 4
mixas84 [53]

Answer:

it's answer is 5 kg

<h2><em>hope </em><em>it </em><em>helps </em><em>you </em></h2>
6 0
3 years ago
Read 2 more answers
Help Me Please!!!!!!!​
iren2701 [21]
B. Conductors allow electricity to go through them only. Such as copper etc.
5 0
3 years ago
Read 2 more answers
A cockroach of mass m lies on the rim of a uniform disk of mass 4.00m that can rotate freely about its center like a merry-go-ro
Alexxandr [17]

Answer:

Explanation:

A. And B. solutions are in the attachment.

C.

When the cockroach walks towards the center of the circle, positive work is done so kinetic energy increases.

5 0
4 years ago
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