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3 years ago

7

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car’s moti

on?

2 answers:

Nitella [24]3 years ago

3 0

The car is staying still. either getting petrol or stopped at a light .

Novosadov [1.4K]3 years ago

3 0

Answer:

The car is not moving

Explanation:

On a position-time graph, the gradient of the line represents the velocity. If the gradient is positive, then the car is speeding up, if it is negative, then the car is slowing down, if it is a horizontal line (where the gradient is zero) then the velocity is zero.

Hence, the horizontal line indicated that the car is not moving.

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An electrician must decide between 3 heaters. The wire that she is using is rated 15A maximum, anything beyond that is a fire ha

lutik1710 [3]

Use the formula: $P=IV$ , where P is power (watts), I is current (amperes) and V is potential difference (volts).

For the first heater:
$900W = I_{1}*120V$
$I_{1} = 7.5A$

For the second heater:
$1800W = I_{2}*120V$
$I_{2} = 15A$

For the third heater:
$3000W = I_{3}*240V$
$I_{3} = 12.5A$

6 0

3 years ago

The heat vaporization for methyl alcohol is 1100 kj/kg. It is 2257 KJ/Kg for water. Thus means that______________.

dusya [7]

Answer: B) Methyl alcohol requires less than half as much energy per kg to evaporate than water does

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 kilo gram of liquid into its vapor state without change in its temperature.

Heat of vaporization is more for water than for methyl alcohol which means more heat is required to convert from liquid to vapour form.

As the Heat of vaporization for methyl alcohol (1100) is almost half as that of Heat of vaporization for water (2257) , it means Methyl alcohol requires less than half as much energy per kg to evaporate than water does.

7 0

2 years ago

The mysterious visitor that appears in the enchanting story The Little Prince was said to come from a planet that "was scarcely

FromTheMoon [43]

Answer:

Part a)

$g = 1 \times 10^{-5} m/s^2$

Part b)

$v = 0.011 m/s$

Explanation:

Part a)

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

$g = \frac{G(\frac{4}{3}\pi R^3 \rho)}{R^2}$

$g = \frac{4}{3}\pi \rho G R$

now we know that

$\rho = \frac{M}{\frac{4}{3}\pi R^3}$

$\rho = \frac{5.98 \times 10^{24}}{\frac{4}{3}\pi (6.37 \times 10^6)^3}$

$\rho = 5523.2 kg/m^3$

now we have

$g = \frac{4}{3}\pi (5523.2) (6.67 \times 10^{-11})(6.5)$

$g = 1 \times 10^{-5} m/s^2$

Part b)

As we know that the escape speed is given as

$v = \sqrt{\frac{2GM}{R}}$

$v = \sqrt{2gR}$

$v = \sqrt{2(1\times 10^{-5})(6.5)}$

$v = 0.011 m/s$

8 0

2 years ago

A body of mass 250g is release from the top of a building. if the body hits the ground with a velocity of 4m/s, calculate th hei

Nutka1998 [239]

$\purple{\maltese}\large\underline{\underline{\sf\:\: Given :}} \\$

» Mass ( m ) of the body = 250 g

» final velocity of the body = 4 m/s

$\\\purple{\maltese}\large\underline{\underline{\sf\:\: To \: \: Find :}} \\$

» The height of the Building= ??

$\\\purple{\maltese}\large\underline{\underline{\sf\:\: Solution :}} \\ \\$

★ <u>Height of the </u><u>building </u><u>by the principal of conservatio</u>n;

$\\\\\begin{gathered}\purple{\maltese}\large\underline{\underline{\sf\:\: using \: formula :}} \\ \\\end{gathered}$

$\bigstar \: \underline{ \boxed{\sf { \pink{ \: \frac{1}{2} \: \: mv { }^{2} = \: mgh \: }}}}$

$\\ \: \: \underline{\textsf {Putting values \: in \: the \: \: given \: formula :}} \\$

$\\ \sf \implies \: \frac{1}{2} \: \: mv { }^{2} = \: mgh \: \\$

$\\ \sf \implies \: \frac{v {}^{2} }{2} \: \: = \: gh \: \\$

$\\ \sf \implies \: h \: = \frac{v {}^{2} }{2g} \: \: \: \\$

$\\ \sf \implies \: h \: = \frac{16 }{20} \: \: \: \\$

$\\ \sf \implies \: h \: = 0.8 \: \: \: m \\$

$\\ \sf \implies \: h \: = 80 \: \: \: cm \\$

henceforth , The height of the Building is <u>0.8 m or 80 cm</u> ..!!!

5 0

1 year ago

A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the

Svetllana [295]

Answer:

a) $\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$ , b) $\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$ , c) $\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

Explanation:

a) The net flux through the cube is:

$\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}$

b) The flux through the right face is:

$\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}$

c) The flux through the left face is:

$\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}$

$\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}$

5 0

3 years ago

Read 2 more answers