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3 years ago

HELP PLS MARKING BRANLIST 100 Pts TAKING TEST RN

Physics

2 answers:

FromTheMoon [43]3 years ago

6 0

Answer:

A. 1.90

I divided 63 by 60 and got 1.05

I multiplied 1.05 by 2 (120 seconds = 2 minutes) and I got 2.1

1.90 is the closest answer.

Bumek [7]3 years ago

5 0

Answer:

It will travel 7,560 m

Explanation:

63 x 120 = 7,560 m

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A) A 5.75 mL sample of mercury has a measured mass of 77.05 g. The density is ___________.

Kitty [74]

Answer:

Density of the sample will be 13.4 kg/L

Explanation:

We have given volume of the sample $V=5.75mL=5.75\times 10^{-3}L$

Mass of the sample $M=77.05gram =77.05\times 10^{-3}kg$

We have to find the density of the sample

Density of the sample is given by

$Density=\frac{mass}{volume}=\frac{77.05\times 10^{-3}}{5.75\times 10^{-3}}=13.4kg/L$

So density of the sample will be 13.4 kg/L

4 0

4 years ago

If the acceleration of an object is zero at some instant in time, what can be said about its velocity at that time? 1. It is neg

Mkey [24]

3. It is not changing at that time

Explanation:

If the acceleration of a body is zero at some instant in time, it implies that the velocity is not changing at that point in time. Velocity is the rate of change of displacement with time.

✓Acceleration and velocity shares a very close relationship.

✓ For a body to accelerate, the velocity must change. Acceleration is defined as the rate of change of velocity with time.

✓If at any point, a body moves with constant velocity i.e the velocity does not change with time, the acceleration becomes zero.

✓ For acceleration to occur, a body must change velocity.

Learn more:

Acceleration brainly.com/question/6323625

#learnwithBrainly

5 0

3 years ago

Give some example acostic using "ELEMENTS"

Jet001 [13]

E=ERBIUM
L=LITHIUM
E=EINSTEINIUM
M=MAGNESIUM
E=EUROPIUM
N=NITROGEN
T=TITANIUM
S=SULPHURDIOXIDE

5 0

3 years ago

A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p

miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}$

$V_{1}=9.18\ V$

The potential on the second plate

$V_{2}=V-V_{1}$

$V_{2}=51.5 -9.18$

$V_{2}=42.32\ v$

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

$C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}$

$C=6.99\times10^{-7}\ F$

$C=0.69\ \mu F$

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0

3 years ago

How can i express 2.99792458 x10^8 in three insignificante figures

Vika [28.1K]

The answer: 3.00 * 10 ⁸ .
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4 0

4 years ago