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2 years ago

8

A hydraulic lift raises 1140 kg car through a height of 2.4 m. What is the potential energy

1 answer:

Sladkaya [172]2 years ago

5 0

1140x9.8x2.4= 26,812.8 significant figures Make it 27,000

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You are in a contest with your friends to see who can drive a golf ball the farthest. should you hit a "line drive" (low to the

WITCHER [35]

Line drive would be more farthest from the very high angle shot because when we increase the angle of flight the range started to increase and at some point it becomes maximum and that angle is 45° after that as you go on increasing the angle it won't be covering more distance as compared to the max at (45°) .

3 0

3 years ago

An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple

Anuta_ua [19.1K]

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of $h=2.3 m$

We need to find the impact speed of apple which can be given by using

$v^2-u^2=2gh$

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

$v^2-0=2\times 9.8\times 2.3$

$v=\sqrt{2\times 9.8\times 2.3}$

$v=6.71\ m/s$

5 0

3 years ago

A closed flat loop conductor with radius 2mm is located in a changing uniform magnetic field. If the emf induced in the loop is

BARSIC [14]

Answer:

Rate of magnetic field is 15923.668 T/sec

Explanation:

We have given radius of loop conductor r = 2 mm = 0.002 m

Induced emf = 2 volt

As the magnetic field and plane are perpendicular to each other so angle between magnetic field and area is 0°

Cross sectional area of the conductor is equal to $A=\pi r^2$

$A=3.14\times 0.002^2=1.256\times 10^{-6}m^2$

Induced emf is given by $e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}$

$2=1.256\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=159235.668T/sec$

6 0

3 years ago

A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amoun

quester [9]

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

$\Delta L = \dfrac{PL}{AE_{Al}}$

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium $E_{Al}$ is:

$E_{Al}= 7*10^{10} \Nm^{-2}$

$\Delta L = \dfrac{PL}{AE_{Al}}$

$\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}$

ΔL = 4.757 × 10⁻⁴ m

ΔL = 0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

5 0

3 years ago

A long, straight, vertical wire carries a current upward. Due east of this wire, in what direction does the magnetic field point

AlexFokin [52]

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

5 0

3 years ago