The diagram shows a trace of a wave on a cathode-ray oscilloscope.

If the torque required to loosen a nut holding a wheel on a car is 48 N · m, what force must be exerted at the end of a 0.23 m l

Andreas93 [3]

Answer:

F = 318.1 N

Explanation:

As we know that torque to open the nut is given by formula

$\tau = \vec r \times \vec F$

so we can write it as

$\tau = rFsin\theta$

now we know that

$\tau = 48 Nm$

r = 0.23 m

angle between force and the wrench is 41 degree

so we have

$48 = (0.23)Fsin41$

$F = \frac{48}{(0.23)sin41}$

$F = 318.1 N$

6 0

3 years ago

Use the periodic table and your knowledge of isotopes to complete these statements.

klasskru [66]

Use the periodic table and your knowledge of isotopes to complete these statements.

When polonium-210 emits an alpha particle, the child isotope has an atomic mass of ⇒ 206.

I-131 undergoes beta-minus decay. The chemical symbol for the new element is ⇒ Xe.

Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of ⇒ 18.

7 0

3 years ago

Read 2 more answers

A golf ball (mass 0.045 kg) is hit with a club from a tee. The plot shows the force on the ball as a function of time in millise

almond37 [142]

Answer:

16.3 m/s

Explanation:

We are given that

Mass of golf ball=m=0.045 kg

We have to find the speed right after the hit.

$b_1=(16-1)\times 10^{-3}=15\times 10^{-3} s$

$b_2=(12-6)\times 10^{-3}=6\times 10^{-3} s$

Force,F=70 N

Change in momentum=Area of trapezium

Area of trapezium,A= $\frac{1}{2}(b_1+b_2) h$

Using the formula

$m(v-u)=\frac{1}{2}(15+6)\times 10^{-3}\times 70$

$0.045(v-u)=\frac{1}{2}(21)\times 10^{-3}\times 70$

$v-u=\frac{1}{2}\times \frac{1}{0.045}\times 21\times 10^{-3}\times 70$

$v-u=16.3$

Initial velocity of golf ball=u=0

$v-0=16.3$

$v=16.3 m/s$

5 0

3 years ago

In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two

IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

$P_{i1}+P_{i2}=P_{f1}+P_{f2}$

where,

$P_{i1}$ = initial momentum of the cue ball = 0.23 N.s

$P_{i2}$ = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

$P_{f1}$ = final momentum of the cue ball = 0.01 N.s

$P_{f2}$ = final momentum of the eight ball = ?

Therefore,

$0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s$

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0

3 years ago

D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weig

s2008m [1.1K]

Answer:

Wc = 7.84 weight of crown

Ww = 7.84 - 6.86 = .98 weight of water displaced

Density = 7.84 / .98 = 8 crown is 8 X that of water

Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold

4 0

2 years ago