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3 years ago

Which of the following are double-displacement reactions? multiple answers

Physics

2 answers:

Sauron [17]3 years ago

7 0

Answer:

C and D

Explanation:

A: is a simple composition

B: is a single replacement

C: C is a double displacement

D: is a double replacement

Julli [10]3 years ago

5 0

Answer:

The answers are C abd D.

Explanation:

Here,

A no. Is a combination or analysis chemical reaction.

B no.is a single displacement reaction as cl2 goes to zn.

C no. Is double displacement reaction.

D no. Is a double displacement reaction.

Hope it helps.....

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How do you find the voltage of a section of a parallel circuit?

sergiy2304 [10]

Answer: Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source. You can find total resistance in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +.

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The weight of the windsurfer is 700 newtons. Calculate the moment exerted by the windsurfer on the sailboat

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3 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

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3 years ago

Which planet has the greatest mass?

Firlakuza [10]

Answer:

Jupiter

Explanation:

The largest planet in our solar system by far is Jupiter, which beats out all the other planets in both mass and volume. Jupiter's mass is more than 300 times that of Earth, and its diameter, at 140,000 km, is about 11 times Earth's diameter.

Note:

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5 0

3 years ago

Read 2 more answers

a disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and m

padilas [110]

The speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

Further Explanation:

Given:

The angular velocity of the disk is $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The radius of the circular disk is $0.1\,{\text{m}}$ .

Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

$v = r \times \omega$

Here, $v$ is the linear speed, $r$ is the radius of the disk and $\omega$ is the angular speed of the disk.

Substitute $0.1\,{\text{m}}$ for $r$ and $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $\omega$ in above expression.

$\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}$

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

${a_c} = \dfrac{{{v^2}}}{r}$

Substitute $1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $v$ and $0.1\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}$

Thus, the speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords: Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

7 0

3 years ago

Read 2 more answers