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3 years ago

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A softball has a (positive/negative) acceleration when it is thrown. A soft ball has a (positive/negative) acceleration when it

is caught

1 answer:

Nadusha1986 [10]3 years ago

8 0

Answer

A softball has a (negative) acceleration when it is thrown. A soft ball has a (positive) acceleration when it is caughtExplanation:

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A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl

Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

4 0

3 years ago

Read 2 more answers

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

Why are chemical changes considered “unseen”?

jeka57 [31]

<span>Because they occur at an atomic level, changing the actual structure of the thing.

</span>

7 0

3 years ago

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Some earthquake waves, a cat’s purr, and elephant communications consist of ultrasonic waves.

Kobotan [32]

The answer to your question is true.

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3 years ago

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When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi

Nikitich [7]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ .

Comparing two wires,

$R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}$

Dividing the above 2 equations,

$\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{A_2 }{A_1} \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{d_2^2 }{d_1^2} \\$

Since $d_2=2d_1$

The above ratio is

$\frac{R_1}{R_2}=\frac{1.68(10^{-8}) }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264$

We also have,

$\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1$

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0

3 years ago