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3 years ago

1. A way of life that involves little or no physical activity is called a(n)_______

Physics

1 answer:

xenn [34]3 years ago

5 0

Answer:

its either A or D

Explanation:

lazy-.

1.Unwilling to do work or make an effort; disinclined to exertion.

2.Causing idleness; relaxed or leisurely.

3.Sluggish; slow-moving.

Sedentary

1.Not moving; relatively still; staying in the vicinity.

2.medicine, of a job, lifestyle, etc. Not moving much; sitting around.

3.obsolete inactive; motionless; sluggish; tranquil

i think it's sedentary

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Use the table to answer the question.

QveST [7]

Answer:

Mechanical Waves

Explanation:

I took the practice

6 0

2 years ago

Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each w

leva [86]

Answer:

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

5 0

3 years ago

Read 2 more answers

A parallel-plate vacuum capacitor has 8.60 J of energy stored in it. The separation between the plates is 3.80 mm . If the separ

creativ13 [48]

Answer:

Part a)

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

$E = 2.62(8.60) = 22.5 J$

Explanation:

As we know that the energy of capacitor when it is not connected to potential source is given as

$U = \frac{Q^2}{2C}$

As we know that initial energy is given as

$8.60 = \frac{Q^2}{2C}$

now we know that capacitance of parallel plate capacitor is given as

$C = \frac{\epsilon_0A}{d}$

now the new capacitance when distance is changed from 3.80 mm to 1.45 mm

$C' = \frac{Cd}{d'}$

$C' = \frac{C(3.80)}{1.45}$

$C' = 2.62 C$

Now the new energy of the capacitor is given as

$E = \frac{Q^2}{2(2.62C)}$

$E = \frac{8.60}{2.62} = 3.28 J$

Part b)

Now if the voltage difference between the plates of capacitor is given constant

now the energy energy of capacitor is

$U = \frac{1}{2}CV^2$

$8.60 = \frac{1}{2}CV^2$

now when capacitance is changed to new value then new energy is given as

$E = \frac{1}{2}C'V^2$

$E = \frac{1}{2}(2.62C)V^2$

$E = 2.62(8.60) = 22.5 J$

6 0

3 years ago

6. Mass affects how fast an object falls.<br> True<br> False

dimulka [17.4K]

The statement: Mass affects how fast an object falls is true.

6 0

3 years ago

Read 2 more answers

The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 10.0 m

adell [148]

Answer:

5.565 V

Explanation:

Radius of coil of generator=r=0.14 m

Length of wire=l=10 m

Magnetic field,B=0.24 T

Angular speed, $\omega=34rad/s$

We have to find the peak emf of the generator.

$N=\frac{l}{2\pi r}=\frac{10}{2\pi\times 0.14}=11$

$A=\pi r^2=\pi (0.14)^2=0.062m^2$

Peak(maximum) induced emf of generator= $E_{max}=NBA\omega$

Using the formula

$E_{max}=11\times 0.24\times 0.062\times 34$

$E_{max}=5.565 V$

3 0

3 years ago