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1 year ago

. A certain sample of metal has a mass of 3 kg and a volume of 250 cm3. What is its density?

Physics

1 answer:

LekaFEV [45]1 year ago

4 0

Given:

The mass of the sample is,

$m=3\text{ kg}$

The volume is

$V=250\text{ cm}^3$

To find:

The density

Explanation:

The volume is,

$\begin{gathered} V=250\text{ cm}^3 \\ =250\times10^{-6}\text{ m}^3 \end{gathered}$

The density is,

$\begin{gathered} d=\frac{m}{V} \\ =\frac{3}{250\times10^{-6}} \\ =12000\text{ kg/m}^3 \\ =12\times10^3\text{ kg/m}^3 \end{gathered}$

Hence, the density is

$12\times10^3\text{ kg/m}^3$

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Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

3 years ago

1. Which of the following states would likely get hit by both hurricanes and tornadoes? (Points : 1)

bogdanovich [222]

I would have to say Texas because, obviously, its on the coast, and because I know for a fact Oklahoma is VERY prone to Tornadoes and I also know Dallas (and surrounding areas) has a few tornadoes a year:)

I hope I helped:)

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5 0

3 years ago

Read 2 more answers

Perform the following calculations and give your answer with the correct number of significant figures

love history [14]

Answer:

see below

Explanation:

a. 0.1886 x 12 =2.2632

This has 2 sig figures so the answer can only have 2 sig figures

2.3

b. 2.995 - 0.16685 =2.82815

The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place

2.828

c. 910 x 0.18945=172.3995

The least number of significant figures is 3 so the answer can only have 3 significant figures

172

3 0

3 years ago

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

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2 years ago

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

4 years ago