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Ainat [17]
3 years ago
10

20 points!

Physics
1 answer:
MariettaO [177]3 years ago
8 0
Do they give answer choices? or is it free write? i’ll help if you tell me!!
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A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally
Alex787 [66]

Answer:

47 ms

well the time it takes to fall 100m is the same time it takes to travel 65m horizontally.

The time to fall vertically, t is sqrt(2d/g). [this comes from d = 1/2at^2]

so t = sqrt(2*100/9.8) = 4.52s.

The vertical speed at that time is g*t = 9.8*4.52 = 44.3m/s

The horizontal speed is the horizontal distance over the same 4.52s, = 65/4.52 = 14.4m/s.

so the final velocity is = sqrt(44.3^2 + 14.4^2) = 47.ms

4 0
3 years ago
A long, cylindrical solenoid with 100 turns per centimeter has a radius of 14.7 cm. If the current through the solenoid changes
oee [108]

Answer:

Explanation:

Given:

dI/dt = 6.21 A/s

n = N/l

= 100 turns/cm

= 100 turns/cm × 100 cm/1 m

Radius, r = 14.7 cm

= 0.147 m

Inductance, L = uo × n^2 × A × l

L/l = 4pi × 10^-7 × (100 × 100)^2 × pi × 0.147^2

= 8.53 H

Emf, E = L × dI/dt

E/l = L/l × dI/dt

= 8.53 × 6.21

= 52.98 V/m

=

4 0
3 years ago
Conductors an insulators
sladkih [1.3K]

Conductors transfer energy (metal is a good one) and insulators stop the transfer of energy (such as rubber or plastic). Sorry if that is wrong, but I hope it helps!

6 0
3 years ago
In order for water to condense on an object, the temperature of the object must be ______ the dewpoint temperature.
enyata [817]

Answer:

at   ( or below)

Explanation:

at the dewpoint.......water will condense out of the air onto the surface

7 0
2 years ago
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
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