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3 years ago

758 j of heat are added to 0.750 kg of copper. how much does its temperature change?(unit=degrees c) PLEASE HELPPPPPPP MEEEEEE

Physics

1 answer:

DiKsa [7]3 years ago

5 0

Answer:

$2.6^{\circ}C$

Explanation:

When a substance is supplied with a certain amount of heat energy, the temperature of the substance increases according to the equation

$Q=mC\Delta T$

where

m is the mass of the substance

Q is the amount of energy supplied

C is the specific heat of the substance

$\Delta T$ is the temperature change

In this problem:

Q = 758 J is the energy supplied

m = 0.750 kg is the mass of the sample

$C=385 J/kg^{\circ}C$ is the specific heat of copper

Re-arranging the equation, we can find the increase in temperature:

$\Delta T=\frac{Q}{mC}=\frac{758}{(0.750)(385)}=2.6^{\circ}C$

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3 years ago

A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the

barxatty [35]

Answer:

2.2 meters

Explanation:

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$k$ is Coulomb's constant.

#The electric field, $E$ at radius r is expressed as:

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From i and ii, we have:

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$r=(KE)/Eq$

#Substitute actual values in our equation:

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Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0

3 years ago

Read 2 more answers

An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2

Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

$s_1 + s_2 = 10 + 4$

$s_1 + s_2 = 14\ kg$

taking moment about B

$s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0$

$s_1 \times 0.8 = 6.4$

$s_1 = 8\ N$

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$s_2 = 14 - 8$

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3 years ago

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

= 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

= 0.8

at x= 5 Km

dy/dx = 0.8(5)

= 4

$p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }$

now insert the values,

$p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km$

→ Now the normal component of acceleration is given by

$a_{n} = \frac{v^{2} }{p}$

= (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

$a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}$

$a = [(0.8)^{2} + (0.457)^{2}]^{0.5}$

a = 0.921 m/s²

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3 years ago

Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh

ahrayia [7]

Answer:

i) FALSE, ii) TRUE, iii) FALSE, iv) FALSE

Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

Let's apply these equations to our case where

nₓ <n_y

i) The expression of the refractive index

nₓ < n_y

$\frac{c}{v_x} = \frac{c}{n_y}$

v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

the statement of FALSE

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2 years ago