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2 years ago

Rpreting Heating Curves

Physics

1 answer:

Sedaia [141]2 years ago

7 0

Answer:

the answer is E-F

Explanation:

why is because the others are lines but E-F is curve

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A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti

allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q = −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0

3 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

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3 years ago

A force of 2 kN is applied to an object to make it move 3.6 m in the direction of the force. Select the correct value of work do

vaieri [72.5K]

Answer:

W= F × d

W= 2kn × 3.6

W= 7.2 J

Work is measured in Joules!

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3 years ago

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c

Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-5}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{5}$

v = -10 cm

The magnification factor of lens

$m=\frac{-10}{-5}$

m = 2

$m=\frac{h'}{h}$

$h'=h\times m$

$h'=3\times 2$

h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

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3 years ago

Does the world actually end on April 21, 2021?

Setler [38]

Answer:

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Explanation:

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3 years ago