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adell [148]
3 years ago
12

Physics question i appreciate your help please

Physics
1 answer:
den301095 [7]3 years ago
4 0

Solution: From the given question, we shall find the vector quantity among the

(A) Time , (B) Velocity, (C) Distance , (D) Speed

Concept: <u>Vector Quantity: </u>All those physical quantities which have magnitude as well as specific directions, are called Vector Quantities.

Here, Time, Distance and Speed have only magnitude but have no directions so they will be scalar quantities.

Now, <u>Velocity:</u> It is defined as the change in displacement per unit time. Since the change in the displacement will be in particular direction only. Hence, velocity will be the vector quantity.

Hence, the option (B) Velocity will be the correct option.

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SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi
tino4ka555 [31]

Answer:

 i = 0.477 10⁴ B

the current flows in the  counterclockwise

Explanation:

For this exercise let's use the Ampere law

                    ∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

        L₀ = (y2-y1)

        L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

        L₁ = 3-0

        L₁ = 3cm

       B L = μ₀ I

       B 2L = μ₀ I

        i = 2 L B /μ₀

        i= 2 0.03 / 4π 10⁻⁷   B

        i = 4.77 10⁴  B

The current is perpendicular to the magnetic field whereby the current flows in the  counterclockwise

8 0
3 years ago
A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo
DIA [1.3K]

Answer:

74^{\circ} C

Explanation:

We are given that

Mass of glass,m=300 g

T_1=23^{\circ}

Volume,V=236cm^3

Mass of water=density\times volume=1\times 236=236 g

Density of water=1g/cm^3

Temperature of hot water,T=87^{\circ}

Specific heat of glass,C_g=0.2cal/g^{\circ}C

Specific heat of water,C_w=1 cal/g^{\circ}C

Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)

Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)

Q_{glass}+Q_{water}=0

300\times 0.2(T_f-23)+236\times 1(T_f-87)

60T_f-1380+236T_f-20532=0

296T_f=20532+1380=21912

T_f=\frac{21912}{296}=74^{\circ} C

5 0
2 years ago
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