6) The temperature of 10 kg of a substance rises by 55°C when heated. Calculate the

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

6 0

3 years ago

A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

Lina20 [59]

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop $A=\pi r^2$

$A=3.14\times 0.045^2=0.00635m^2$

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field $dB=250-350=100mT$

Emf induced is given by

$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V$

Magnitude of induced emf is equal to 0.00635 V

7 0

3 years ago

A change in position of an object relating to time

MaRussiya [10]

The answer is: Motion!

Have a great day

3 0

3 years ago

Read 2 more answers

Calculate the power of convex lens of focal length 20cm

maxonik [38]

D = 1/f, where D is the power in diopters and f is the focal length in meters.

D=1/20

<u>D=0.05</u>

6 0

4 years ago

The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26

Delvig [45]

Answer:

(a) The magnetic energy density in the field is 6.366 J/m³

(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

magnitude of magnetic field inside solenoid, B = 4 T

inner diameter of solenoid, d = 6.2 cm

inner radius of the solenoid, r = 3.1 cm = 0.031 m

length of solenoid, L = 26 cm = 0.26 m

(a) The magnetic energy density in the field is given by;

$u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3$

(b) The energy stored in the magnetic field within the solenoid

$U_B = u_B V\\\\U_B = u_B AL$

$U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\$

6 0

3 years ago