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1 year ago

Which of these is NOT a physical property?

1 answer:

8 0

I believe the answer is chemical reactivity because: Characteristics such as melting point, boiling point, density, solubility, color, odor, etc. are physical properties.

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Which of the following is not the one of the influencing factors when it comes to body type

serg [7]

Where are the answer choice ?

3 0

2 years ago

Read 2 more answers

A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi

S_A_V [24]

Answer:

10.2 Watt

Explanation:

$N$ = number of turns in flat coil = 160

$A$ = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

$B$ = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

$t$ = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

$E = \frac{- N A \Delta B}{t}$

$E = \frac{- (160) (0.20) (- 0.80)}{2}$

$E$ = 12.8 volts

$R$ = Resistance of the coil = 16 Ω

Power is given as

$P = \frac{E^{2}}{R}$

$P = \frac{(12.8)^{2}}{16}$

$P$ = 10.2 Watt

6 0

2 years ago

A 1000 kg elevator is accelerated upward at a rate of 0.70 m/s2. What is the tension in the cable pulling the elevator upward wh

Masteriza [31]

Answer:

Tension is also known as Force...

and Force is mass× acceleration.

so....1000×0.70=700N

5 0

3 years ago

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A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg

motikmotik

Answer:

$\omega_{f}=1.634\ rad/s$

Explanation:

given,

diameter of merry - go - round = 2.40 m

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg

initial angular momentum of the system

$L_i = I\omega_i$

$L_i =356\times 1.80$

$L_i =640.8\ kg.m^2/s$

final angular momentum of the system

$L_f = (I_{disk}+mR^2)\omega_{f}$

$L_f = (356 + 25\times 1.2^2)\omega_{f}$

$L_f= (392)\omega_{f}$

from conservation of angular momentum

$L_i = L_f$

$640.8= (392)\omega_{f}$

$\omega_{f}=1.634\ rad/s$

8 0

2 years ago

Read 2 more answers

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

8 0

2 years ago