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3 years ago

How many Newtons of force does it take to move a 53 kilogram object?

Physics

1 answer:

Sidana [21]3 years ago

3 0

519.75N

explanation: good luck on whatever assignment

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A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fir

bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.

So with y(1) = v - 4.9 = 0 we have

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0

3 years ago

ASAP! AWARDING BRAINLIEST see attached photo.

stellarik [79]

Answer:d

Explanation:

6 0

3 years ago

Read 2 more answers

True or False: Every lightning strike produces thunder.

amid [387]

False,

True,

A build up of positive charge builds up on the ground under the cloud, attracted to the negative charge in the cloud. The earth's positive charge concentrates to anything that sticks up - trees, lightning conductors,

100

8 0

3 years ago

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

3 years ago

Saki is ice-staking to the left with velocity of 5 m/s. A steady breeze starts blowing. Causing saki to accelerate leftward at a

Pavlova-9 [17]

Answer:

Final velocity will be equal to 14 m/sec

Explanation:

We have given initial velocity u = 5 m/sec

Constant acceleration is given $a=1.5m/sec^2$

Time t = 6 sec

We have to find the final velocity

From first equation of motion $v=u+at$ , here v is final velocity, u is initial velocity , a is acceleration and t is time

So $v=5+1.5\times 6=14m/sec$

So equal final velocity will be equal to 14 m/sec

4 0

3 years ago