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3 years ago

Manuel is holding a 5kg box. How much force is the box exerting on him?In what direction

1 answer:

3 0

The force the box is exerting on Manuel is the weight of the box, downward:
$W=mg=(5 kg)(9.81 m/s^2)=49.05 N$
and this force is perfectly balanced by the constraint reaction applied by Manuel's hand, pushing upward.

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I'd like you to think back on 2 events in

strojnjashka [21]

Answer:

Holi

biwali

these are best events and love to celebrate withmy family and friends it contains lot of happiness and joy

7 0

2 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm

Karo-lina-s [1.5K]

Answer:

the thermistor temperature = $325.68 \ ^0 \ C$

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

$T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K$

Resistance of the thermistor $R_1 =$ 20,000 ohms

Material constant $\beta$ = 3650

Resistance of the thermistor $R_2$ = 500 ohms

Using the equation :

$R_1 = R_2 \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{R_1}{ R_2} = \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

Taking log of both sides

$In \ \frac{R_1}{ R_2} = In \ \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{T_1} - \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}$

${T_2} = \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}$

Replacing our values into the above equation :

${T_2} = \frac{3650*373}{3650 - In (\frac{20000}{500})373}$

${T_2} = \frac{1361450}{3650 - 3.6888*373}$

${T_2} = \frac{1361450}{3650 - 1375.92}$

${T_2} = \frac{1361450}{2274.08}$

${T_2} = 598.68 \ K$

${T_2} = 325.68 \ ^0 \ C$

Thus, the thermistor temperature = $325.68 \ ^0 \ C$

4 0

3 years ago

Which is one<br> step that geologists use to find the epicenter of an earthquake?

Kisachek [45]

Answer:

triangulation

Explanation:

using your compass

5 0

3 years ago

Read 2 more answers

An Argon laser (λ = 5.0×102nm) shines down a silica glass fiber-optic cable with index of refraction 1.46. What is the speed of

Yanka [14]

Answer:

The speed of the laser light in the cable, $c_f=2.1\times 10^8\ m/s$

Explanation:

It is given that,

Wavelength of Argon laser, $\lambda=5\times 10^2\ nm=5\times 10^{-7}\ m$

Refractive index, n = 1.46

Let $c_f$ is the speed of the laser light in the cable. The speed of light in a medium is given by :

$c_f=\dfrac{c}{n}$

$c_f=\dfrac{3\times 10^8\ m/s}{1.46}$

$c_f=2.05\times 10^8\ m/s$

$c_f=2.1\times 10^8\ m/s$

So, the speed of the laser light is $2.1\times 10^8\ m/s$ . Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers