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timama [110]
3 years ago
9

Given a force of 88N and an acceleration of 4m/s2 what is the mass

Physics
1 answer:
anygoal [31]3 years ago
6 0
Use newtons second law F=ma, plug in the given values which gives us the answer of 22 kg for the mass
You might be interested in
A scientific hypothesis starts with the lowest level of acceptance but can become stronger as they are used and survive repeated
Pachacha [2.7K]

Answer:

True.

Explanation:

Hypotheses are supposed to be tested, to be proven correct. They usually start from a small thought without 100% correct, but as you do further testing, they become stronger.

3 0
3 years ago
Read 2 more answers
Select all of the statements that are true.
natta225 [31]

The correct statements are "Each orbit holds a fixed number of electrons" and "The n=1 orbit can only hold two electrons." According to the Bohr model, the maximum number of electrons that can occupy an orbit is given by 2n^2, where n is the number of the orbit. For instance, when n=1 it means 2n^2= 2(1)^2=2. This particular orbit can only hold up to two electrons. Even though the electrons can gain energy and move to higher orbits or electrons from higher orbits can lose energy and drop to the n=1 level, the energy level would not allow more electrons to enter the orbit once it is full. Again the octet rule, which states that atoms achieve stability by having 8 valence electrons, limits the maximum number of electrons that can be occupied by an orbit. The gain and loss of electrons is done to achieve the noble gas configuration and once that is reached no more electron can be added to an orbit

8 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
How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______
lukranit [14]

Answer:

A. 8.64 secs.

B. 14.58 secs.

C. 26.002 secs.

D. 33.46secs.

Explanation:

A. P wave would travel 7000km

p-wave travels on a speed of 13.5km/s

= 7000km/13.5km/s

= 8.64 secs.

B. S-wave time to travel 7000km

s-wave travels on a speed of 8km/s

= 7000km/8km/s

= 14.58 secs.

C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).

= 7000km to miles

= 4349.598m/2.788m/s

= 26.002 secs.

D. Rayleigh wave to travel 7,000 km

10,000m/s ( 2.1667 m/s ).

= 7000km to miles

= 4349.598m/2.1667m/s

= 33.46secs.

5 0
4 years ago
A car is driving due South<br> counterclockwise on a<br> circular track.
Alex Ar [27]

Answer:

a

Explanation:hope this helps

3 0
3 years ago
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