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8_murik_8 [283]
3 years ago
12

Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb

er: cal=m*deltaT
(WRITE THE WORK!/Give explanation.)
Physics
1 answer:
GuDViN [60]3 years ago
7 0

Answer:

Q = 12540  J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

Q=mc\Delta T

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

Q=50\times 4.18\times (60-0)\\Q=12540\ J

So, 12540 J of energy is used to heat the water.

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A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring
azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 \frac{N}{m}

Angle \theta = 34°

From the free body diagram

Force acting on the box = mg sin\theta

⇒ F = 3 × 9.81 × \sin34

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

F_{sp} = k x

F_{sp} = 120 x -------- (2)

The net force acting on the body is given by

F_{net} = ma

Since acceleration of the box is zero so

F_{net} = 0

F - F_{sp} = 0

F = F_{sp}

Put the values from equation (1) & (2) we get

16.45 = 120x

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0
3 years ago
A small car with mass of 0.800 kg travels at a constant speed
Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

7 0
2 years ago
If the average velocity during the athlete's walk back
goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

4 0
3 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
kramer

Explanation:

The gravitational force equation is the following:

F_G = G * \frac{m_1 m_2}{r^2} \\

Where:

G = Gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0
3 years ago
What is the IMA of the following pulley system?<br><br>34567
Lynna [10]

Answer:

    IMA of given system =   \frac{F_{r} }{F_{e} }

Explanation:

  • The "Ideal Mechanical advantage" (IMA) of given pulley is \frac{F_{r} }{F_{e} } .
  • Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by F_{r}. The input force is analogous or equivalent to the effort applied i.e. F_{e} .
  • Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is  \frac{F_{r} }{F_{e} } .
  • Its mathematical reference would be:

                                                IMA =   \frac{F_{r} }{F_{e} }

6 0
3 years ago
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