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3 years ago

How much heat (in calories) are needed to raise the temperature of 60 g of water

Physics

1 answer:

laila [671]3 years ago

3 0

Answer:

Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius.

Explanation:

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If a race car wants to change its velocity then a net force is<br> required to cause the necessary

pashok25 [27]

Answer:acceleration

Explanation:

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4 years ago

A piece of indium with a mass of 16.6 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to

blagie [28]

Answer:

the density of indium is 7.2 g/cm^3

Explanation:

The computation of the density of indium is shown below:

Given that

Mass = 16.6 g

Volume = 48.6 c,^3 - 46.3cm^3 = 2.3 cm^3

Based on the above information

As we know that

Density = mass ÷ volume

So,

= 16.6g ÷ 2.3 cm^3

= 7.2 g/cm^3

hence, the density of indium is 7.2 g/cm^3

We simply applied the above formula so that the correct value could come

And, the same is to be considered

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3 years ago

You set a small eraser at a distance of 0.27 m from the center of a turntable you find in the attic and establish that when the

Licemer1 [7]

Answer with Explanation:

We are given that

Distance,r=0.27 m

Tangential speed=v=0.49 m/s

a.Angular speed , $\omega=\frac{v}{r}$

Using the formula

$\omega=\frac{0.49}{0.27}=1.8 rad/s$

$\omega=\frac{2\pi}{T}$

$T=\frac{2\pi}{\omega}$

Time period, $T=\frac{2\pi}{1.8}=3.49 s$

b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m

c.Maximum speed, $V_{max}=0.49 m/s$

d.Maximum acceleration= $a=r\omega^2=0.27(1.8)^2=0.87 m/s^2$

5 0

4 years ago

Read 2 more answers

Changes in the forms of energy are called

Dima020 [189]

<span>Changes in the forms of energy are called "Energy Conversions"

Hope this helps!</span>

5 0

3 years ago

A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for

Elena-2011 [213]

Answer:

1) 10.1 s 2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

In the vertical direction, as the bomb is dropped, its initial velocity is 0.
So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

$\Delta y = \frac{1}{2}*a*t^{2} (1)$

where Δy = -500 m (taking the upward direction as positive).
a=-g=-9.8 m/s²
Replacing these values in (1), and solving for t, we have:

$t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s$

The time required for the bomb to reach the earth is 10.1 s.

In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

$x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.$

The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

$v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s$

If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
So, when the bomb hits the ground, the helicopter will be exactly over it.

8 0

4 years ago