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3 years ago

There is a puncture in the inner tor of your bicycle. Give a method to detect the position of the puncture

Physics

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer:

Explanation:

The first method to engage is to listen to where the sound of air in the inner Tor escaping originated and look to see if u can find it. You can then feel the escape air with your hand.

You can Put it inside a container of water and see the bubble and rotate the inner tube to pass all of it through the water

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If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

6 0

3 years ago

Read 2 more answers

What is the weight of a dust particle

Alexandra [31]

Answer:

The mass of a dust particle is 7.53 x 10^-10 kg. (Hope it helped.)

Explanation:

3 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

You want to lean your dad's ladder on a smooth wall. If the mass of ladder is 4.42 kg and coefficient

iren [92.7K]

Answer:

angle minimum θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

N - W = 0

N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

the friction force formula is

fr = μ N

fr = μ W

we substitute

μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0

μ sin θ - cos θ + ½ cos θ= 0

μ sin θ - ½ cos θ = 0

sin θ / cos θ = 1/2 μ

tan θ = 1/2 μ

θ = tan⁻¹ (1 / 2μ)

θ = tan⁻¹ (1 (2 0.57))

θ = 41.3º

7 0

3 years ago

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Alex777 [14]

The resultant speed of the plane is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Hence, we have that the resultant speed of the plane is (3) 226 m/s

Have a nice day!

5 0

3 years ago