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alexdok [17]
3 years ago
15

Mass of block A above is 25-kg. Calculate the force of gravity on block A.

Physics
1 answer:
stiv31 [10]3 years ago
4 0

Answer:w+mg

Explanation:

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Single Replacement: HCI + Zn
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During a single replacement, the Zn and H will switch places. That means the product formed will be ZnCl and H2.

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You notice that the price of Blu-ray players falls and the quantity of Blu-ray players sold increases. This set of observations
satela [25.4K]

Answer:

Option C

Explanation:

When this type of situation occurs where the price falls and the sale of a product increases then a shift is observed in the supply curve towards right as a result of this shift there is a need to shift down the demand curve

Thus When supply curve is shifted to the right the demand curve is moved down accordingly.

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3 years ago
If a car is moving to the left with constantvelocity, one can conclude thatthere mustbe no forces applied to the car.the netforc
Allisa [31]

Answer:

the net force applied to the car is zero.

Explanation:

According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):

a=\frac{F}{m}

where m is the object's mass.

In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is

the net force applied to the car is zero.

3 0
3 years ago
An old-fashioned LP record rotates at 33 1/3 RPM
Ksenya-84 [330]

Answer:

Part a) \frac{5}{9}\ \frac{rev}{sec}

Part b) \frac{9}{5}\ \frac{sec}{rev}

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

33\frac{1}{3}\ \frac{rev}{min}

Convert mixed number to an improper fraction

33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}

Remember that

1\ min=60\ sec

Convert rev/min to rev/sec

\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}

Simplify

\frac{5}{9}\ \frac{rev}{sec}

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

\frac{9}{5}\ \frac{sec}{rev}

4 0
3 years ago
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