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skelet666 [1.2K]

2 years ago

7

N/an/annnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

1 answer:

disa [49]2 years ago

7 0

Answer:

hiii

Can I get the Brainiest?

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I was watching quantum leap and I was wondering if we can actually travel through different people's body without knowing it's a

alukav5142 [94]

Answer:

No you could not do that because if you tried even if you where to go super fast they would feel a breif second of pain before being completely riped from there body

7 0

2 years ago

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

$v_i = 5.10 m/s$

displacement of the package as it is dropped on ground

$d = -105 m$

acceleration is due to gravity

$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

7 0

3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

2 years ago

Read 2 more answers

An electric heater is operated by applying a potential difference of 78.0 V across a nichrome wire of total resistance 7.00 Ω. a

aleksley [76]

PART A)

Here we know that

potential difference across the wire is

$V = 78 volts$

resistance of wire is

$R = 7 ohm$

now by ohm's law

$V = iR$

$78 = i(7 ohm)$

$i = 11.14 A$

Part b)

Power rating is defined as rate of electrical energy

it is defined as

$P = iV$

now we have

$P = 11.14 ( 78)$

$P = 869.1 Watt$

8 0

2 years ago

A student was walking with a mass of 40 kilograms and an acceleration of 1 m/s/s. What is the force exerted on her?

ValentinkaMS [17]

Answer:

B

Explanation:40x1=40 40=b

brainlyest plz

3 0

3 years ago