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adelina 88 [10]
2 years ago
9

Weigt A box sitting still on the ground by itself has a weight of 700N, what is the mass? (gravity's acceleration is 9.80 m/s2)​

Physics
1 answer:
Alex2 years ago
7 0
  • Force=700N
  • Acceleration=9.8m/s^2

Using newtons law

\\ \rm\longmapsto F=ma

\\ \rm\longmapsto m=\dfrac{F}{a}

\\ \rm\longmapsto m=\dfrac{700}{9.8}

\\ \rm\longmapsto m=71.4kg

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A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc
saul85 [17]

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

F = kx

now we know by Newton's II law

F = ma

so here we have

a = \frac{F}{m}

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

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3 years ago
Help!!!, combination circuits, Physics
Kaylis [27]

Current and voltage on each resistor:

I_1 = 3.98 A, V_1 = 3.98 V

I_2=0.015 A, V_2 = 0.075 V

I_3 = 0.4 A, V_3 = 0.4 V

I_4 = 0.385 A, V_4 = 0.77 V

I_5 = 0.585 A, V_5 = 1.17 V

I_6 = 3.01 A, V_6 = 6.02 V

I_7 = 0.97 A, V_7 = 4.85 V

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega

This resistor is in series with resistor 4, so:

R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega

This resistor is in parallel with resistor 5, therefore:

R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega

This resistor is in series with resistor 7, so:

R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega

This resistor is in parallel with resistor 6, so:

R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega

Finally, this combination is in series with resistor 1:

R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A

And we can also find the potential difference across resistor 1:

V_1=I_1 R_1=(3.98)(1.0)=3.98 V

This means that the voltage across resistor 6 is

V_6=V-V_1=10-3.98=6.02 V

And so, the current on resistor 6 is

I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

I_7=I-I_6=3.98-3.01=0.97 A

And so the voltage across resistor 7 is

V_7=I_7 R_7=(0.97)(5.0)=4.85 V

The voltage across resistor 5 is

V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V

And so the current is

I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A

The current through resistor 4 is

I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A

And therefore its voltage is

V_4=I_4 R_4 = (0.385)(2.0)=0.77 V

So, the voltage through resistor 3 is

V_3=V_5-V_4=1.17-0.77=0.4 V

And the current is

I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A

Finally, the current through resistor 2 is

I_2=I_4-I_3=0.5-0.385=0.015 A

And so its voltage is

V_2=I_2R_2=(0.015)(5.0)=0.075 V

Learn more about current and voltage:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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A uniform brick of length 21 m is placed over
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Answer:

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Now let's look at both bricks as a combined mass.  We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m.  And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge.  So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

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3 years ago
A spring is stretched to a displacement of 3.4 m from equilibrium. Then the spring is released and allowed to recoil to a displa
Stells [14]
Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
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3 years ago
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