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3 years ago

14

Identify the simple machine shown below.

1 answer:

Tresset [83]3 years ago

4 0

Answer:

I believe that is A. Inclined plane

Explanation:

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A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?

GalinKa [24]

Answer:

The electric flux through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

Explanation:

Given

$Radius,\ r = 0.03m\\Charge,\ q =7.6\µC$

Required

Find the electric flux

Electric flux is calculated using the following formula;

Ф = q/ε

Where ε is the electric constant permitivitty

ε $= 8.8542 * 10^{-12}$

Substitute ε $= 8.8542 * 10^{-12}$ and $q =7.6\µC$ ; The formula becomes

Ф = $\frac{7.6\µC}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}$

Ф = $\frac{7.6}{8.8542} *10^{12-6}}$

Ф = $0.85834970974 *10^{12-6}}$

Ф = $0.85834970974 *10^{6}}$

Ф = $8.5834970974 *10^{5}}$

Ф = $8.58 *10^{5} \frac{Nm^2}{C}$

Hence, the electric flux through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

7 0

3 years ago

the acceleration due to gravity on earth is 9.80 m/s2. if the mass of a giraffe is 1,470 kg, what is the weight of the giraffe?

Brrunno [24]

Weight = Mass * gravity

= 1470* 9.8 = 14406 N ≈ 14,400 N

5 0

3 years ago

Read 2 more answers

If two cars have the same velocity do they have the same acceleration

Allisa [31]

No, if it takes less time, the acceleration is greater.

Hope this helps!

3 0

3 years ago

If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the 1200.-kg pick-up truck behind him continues tr

Alborosie

Answer: 20.2 m/s

Explanation:

From the question above, we have the following data;

M1 = 800kg

M2 = 1200kg

V1 = 13m/s

V2 = 25m/s

U (common velocity) =?

M1V1 + M2V2 = (M1 + M2). U

(800*13) + (1200*25) = (800+1200) * U

10400 + 30000 = 2000u

40400 = 2000u

U = 40400 / 2000

U = 20.2 m/s

5 0

3 years ago

Read 2 more answers

Heat is extracted from a certain quantity of steam at

vodomira [7]

Answer: $v=2452.91 m/s$

Explanation:

Given

initially steam is at $100^{\circ}C$ and converted to $0^{\circ} C$ ice

Let m be the mass of steam

latent heat of fusion and vaporization for water is

$L_f=3.33\times 10^5 J/kg$

$L_v=2.26\times 10^6 J/kg$

Heat required to convert steam in to water at $100^{\circ}C$

$Q_1=m\times L_v=m\cdot 2.26\times 10^6 J$

Heat required to lower water temperature to $0^{\circ}C$

$Q_2=m\times c\times \Delta T$

$Q_2=m\times 4.184\times (100)$

$Q_2=4.184m\times 10^5 J$

Heat required to convert $0^{\circ}C$ water to ice at $0^{\circ}C$ is

$Q_3=m\times L_f$

$Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J$

$Q=Q_1+Q_2+Q_3$

$Q=(2.26+0.4184+0.33)m\times 10^6 J$

$Q=3.0084m\times 10^6 J$

So this energy is equal to kinetic energy of bullet of mass m moving with velocity v

$Q=\frac{1}{2}mv^2$

$3.0084m\times 10^6=\frac{1}{2}mv^2$

$v^2=3.0084\times 2\times 10^6$

$v=2.452\times 10^3 m/s$

$v=2452.91 m/s$

5 0

4 years ago