Explain how you can change your workout routine using each of the SPORT and FITT principles.

A student measured the volume of three rocks. The diagram above shows the

victus00 [196]

Answer:

option B is right

...............

4 0

2 years ago

Suppose you have measured the diffraction pattern of a grating with d = 0.19 mm and have found that the spots were separated by

cestrela7 [59]

Answer:

d = 0.076 mm

Explanation:

Given data

diffraction pattern d1 = 0.19 mm = 0.019 cm

separated s(1) = 1.8 cm

separated s(2) = 4.5 cm

to find out

d2 for an unknown

solution

we know here that spacing in between the diffraction fringe is always inversely proportional to diffraction grating so

we will apply here formula for unknown d that is

d1 (s1 / L) = d2 (s2 /L)

d2 = d1 × s(1) / s(2)

put here all thes evalue we get d2

d2 = d1 × s(1) / s(2)

d2 = 0.019 × 1.8 / 4.5

d2 = 0.0076 cm

d2 = 0.076 mm

7 0

3 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

A sample of a solid nonmetal is compared to a sample of a solid metal. Both samples have the same volume. What is the most likel

KATRIN_1 [288]

The weight could be different, metals have a higher mass than nonmetals, so when occupying the same amount of space, the weight of the metal is far more.

7 0

3 years ago

Read 2 more answers

A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?

Mkey [24]

The car's mass is 1600 kg.

Its weight is (mass) x (gravity).

On Earth, that's (1600 kg) x (9.8 m/s²) = 15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.

5 0

3 years ago