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4 years ago

6

Two types of specialized fingerprint

1 answer:

Akimi4 [234]4 years ago

7 0

Two types of fingerprints are

.1. arches

2. loopes

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A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

5 0

3 years ago

Is prey abiotic or biotic?

Ray Of Light [21]

Biotic because of the prey of the reasoning

8 0

3 years ago

A plastic tube placed from the ventricles of the brain to the peritoneal cavity in order to continuously remove excess cerebrosp

irina1246 [14]

Answer:

ventriculoperitoneal

Explanation:

A VP or ventriculoperitoneal shunt is a medical equipment or a device which is used to relieve pressure from the brain that is caused by the accumulation of fluid in the brain. Ventriculoperitoneal shunting is a medical procedure that is primarily used to treat a condition known as hydrocephalus. This condition occurs when the more cerebrospinal fluid (CSF) gets collected in the ventricles of the brain than required.

Hence the answer is ---

ventriculoperitoneal

4 0

3 years ago

A raging bull of mass 700kg runs at 10 m/s. How much kinetic energy does it have?

Illusion [34]

Answer: 35000 J or 35kJ

Explanation:

Use equation for kinetic energy : Ek=mV²/2

m=700 kg

V=10m/s

-------------------

Ek= 700kg*100m²7s²/2

Ek=35000 J=35kJ

6 0

4 years ago

A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60∘ away from vertical increases fro

bixtya [17]

Answer:

The induced emf is $|\epsilon|=0.0261 V$

Explanation:

From the question we are told that

The number of turn is $N = 100$

The diameter of the coil is $d = 2.0 cm = 2.0 *10^{-2} m$

The uniform magnetic at initial is $B_i = 0.50 T$

The uniform magnetic at initial is $B_f = 1.50 T$

The time taken is $t = 0.60s$

The angle the magnetic field makes with vertical is $\theta = 60^o$

Generally induced emf is mathematically represented as

$\epsilon = -N \frac{d \o}{dt}$

where $d \o$ is the change magnetic flux

Magnetic flux is mathematically represented as

$\O = \= B \cdot \= A$

$= BA cos \theta$

Substituting this above

$\epsilon = -N \frac{d (BA cos \theta)}{dt}$

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

Where B is the magnetic field and A is the area which is mathematically evaluated as

$A = \frac{\pi d^2}{4}$

Substituting values

$A = \frac{\pi (2.0 *10^{-2})^2}{4}$

$A= 3.142*10^{-4}m^2$

From the equation of emf

$\epsilon = -N A \frac{d (B cos \theta)}{dt}$

dB = $B_2 -B_1$

So

$|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}$

substituting values

$|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}$

$|\epsilon|=0.0261 V$

6 0

3 years ago