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o-na [289]
4 years ago
14

How much heat is needed to raise the temperature of 90g of water by 64oC?

Physics
1 answer:
wel4 years ago
8 0
The heat released or absorbed by a process where temperature changes is calculated from the product of the mass of the sample, its heat capacity (4.18 J/g C for water) and the change of the temperature.

Heat = mC(T)
Heat = 90 (4.8) (64) = 27648 J or 27.65 kJ needed
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Which of of the following is a physical change ice melting into a puddle of water,Leaves changing color in fall, Or baking a cak
Tom [10]

Answer:

ice melting

Explanation:

Because once it melts you can change it back to ice.

4 0
3 years ago
A car is moving at 14 m/s. After 30 s, its speed inncreased to 20 m/s. What is the acceleration over time
Juli2301 [7.4K]

Answer:

0.2 m/s^2

Explanation:

initial speed 14m/s

final speed 20m/s

acceleration:

(20m/s - 14m /s) /30s = (6m/s)/30s = 0.2 m/s^2

7 0
3 years ago
Which best explains satellites? the moon is a satellite of the earth, and the earth is a satellite of the sun. mercury and pluto
Basile [38]
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6 0
3 years ago
Read 2 more answers
A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is
Kruka [31]

Answer:

3.76 m/s

Explanation:

Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.

v' = dx(t)/dt..................... Equation 1

Where v' = instantaneous velocity, x = distance, t = time.

Given the expression,

x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³

x(t) = 28 + 12.4t - 0.0450t³

Differentiating x(t) with respect to t.

dx(t)/dt = 12.4 - 0.135t²

dx(t)/dt = 12.4 - 0.135t²

When t = 8.00 s.

dx(t)/dt = 12.4 - 0.135(8)²

dx(t)/dt = 12.4 - 8.64

dx(t)/dt = 3.76 m/s.

Therefore,

v' = 3.76 m/s.

Hence, the instantaneous velocity = 3.76 m/s

8 0
3 years ago
Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s
ozzi

Answer:

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe       ..... (1)

T Cosθ = mg     .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}

tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}

tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}

tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}

As θ is very small, so tanθ and Sinθ is equal to θ.

\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

7 0
3 years ago
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