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3 years ago

Describe the difference between contact forces and field forces.

1 answer:

8 0

Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.

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What is different about the different parts of the electromagnetic spectrum

Archy [21]

Answer:

Each part of the spectrum has a different range of wavelengths (which correspond to frequency, color, and energy) at which you can find it.

4 0

3 years ago

Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy pl

kogti [31]

Answer:

Part a)

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

Explanation:

As per Newton's II law we know that

F = ma

so we will have

$a = \frac{F}{m}$

so we will have

$a = \frac{F_1 + F_2}{m}$

Part a)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}$

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}$

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}$

$a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}$

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

3 0

3 years ago

A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base

-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g= $426\times 10^{-3} kg$

1 kg=1000 g

Radius,r= $\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

$I=\frac{2}{2}mr^2$

a.By law of conservation of energy

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh$

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy= $\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J

6 0

3 years ago

At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0

2 years ago

If a car is speeding up,its initial speed is ______than its final speed

Karolina [17]

Initial speed is less than final speed

5 0

3 years ago