How many significant figures are in 246.32

A 0.30-kg object is traveling to the right (in the positive direction) with a speed of 3.0 m/s. After a 0.20 s collision, the ob

andre [41]

Answer:

$I = -2.1 kg.m/s$

Explanation:

Given,

mass of the object,m = 0.30 Kg

initial speed, v_i = 3 m/s

time of collision = 0.20 s

final speed, v_f = -4 m/s

Impulse = change in momentum

$I = m (v_f-v_i)$

$I = 0.30\times (-4-3)$

$I = -2.1 kg.m/s$

Hence, impulse of the object is equal to $I = -2.1 kg.m/s$

5 0

2 years ago

A tortoise can run with a speed of 0.14 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, b

Reptile [31]

speed of tortoise is given as v1 = 0.14 m/s

speed of hare is given as v2 = 20*0.14 = 2.8 m/s

now let say the total length of the path is "d"

so the total time taken by the tortoise to cover this

$t = \frac{d}{0.14}$

now given that hare took rest for 1 min

so total time of run for hare is (t - 60)s

so the distance that hare covered is given by

$d - 0.30 = 2.8 * (t - 60)$

now by above two equations

$d - 0.30 = 2.8 * \frac{d}{0.14} - 168$

$168 - 0.30 = (20 - 1)d$

$d = 8.82 m$

and the time t is given by

$t = \frac{8.82}{0.14}$

$t = 63 s$

so part a)

t = 63 s

part b)

d = 8.82 m

4 0

3 years ago

A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an

frozen [14]

Answer:

a) 2.278 x 10^-5 volts

b) 1.139 x 10^-6 Ampere

c) 2.59 x 10^-11 W

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = $\pi r^{2}$

==> A = 3.142 x $0.002^{2}$ = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

$I$ = E/R

where R is the resistor

$I$ = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere

c) power delivered to the resistor is given as

P = $I$ E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W

4 0

2 years ago

A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A. It is placed in a magnetic field of 0.90 T with

Brums [2.3K]

Answer:

3.08 Nm

Explanation:

N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree

The angle made with the normal of the coil, theta = 90 - 30 = 60 degree

Torque = N I A B Sin Theta

Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60

Torque = 3.08 Nm

7 0

3 years ago

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0

2 years ago