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3 years ago

6

FORMULA PLEASE AND explINATION

1 answer:

Alex_Xolod [135]3 years ago

7 0

2 Newtons to the right.

3 newtons are needed to over come the friction. There are 2 left over.

So the answer is 2 newtons to the right.

5 - 3 = 2

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What are the similarities between strong nuclear force and weak nuclear force

Artemon [7]

Alike because they both act on the quarks making up the nucleons and they have very short ranges. The Strong Nuclear Force is an attractive force between protons and neutrons that keep the nucleus together and the Weak Nuclear Force is responsible for the radioactive decay of certain nuclei. Which also makes them very different

7 0

3 years ago

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

3 years ago

A typical student listening attentively to a physics lecture has a heat output of 100 w . how much heat energy does a class of 1

nexus9112 [7]

Since each student emits 100 W, so 170 students will emit:

total heat = 100 W * 170 = 17,000 W

Convert minutes to seconds:

time = 50 min * (60 s / min) = 3000 s

The energy is therefore:

E = 17,000 W * 3000 s

<span>E = 51 x 10^6 J = 51 MJ</span>

7 0

3 years ago

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s. The circle is parallel to the xy pl

Lynna [10]

It's 60.... i have used the formula
L=r^2mw
and didn't use the value 0.75...

6 0

3 years ago

Read 2 more answers

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

3 years ago