When the balloon sticks to the wall (assuming it sticks to the wall). It is

A bucket of water is being raised from a well using a rope. If the bucket of water has a mass of 6.2 kg, how much force (in N) m

Sholpan [36]

Answer:

6.2N force

Explanation:

According to Newton's second law of motion, force is equal to the product of the mass of a body and its acceleration. Mathematically,

Force = mass × acceleration

Given mass of bucket of water = 6.2kg

acceleration of the bucket = 1m/s²

Force exerted on the rope = 6.2 × 1

= 6.2N

5 0

3 years ago

PLS ANSWER FAST WILL GIVE BRAINLEST!!

egoroff_w [7]

Answer:

from

force =mass x acceleration

mass = force/acceleration

m = f/a

m = 7.5/15

m=0.5kg

3 0

3 years ago

Read 2 more answers

Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol

OlgaM077 [116]

Answer:

481.76 J/mol

133.33 K

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

Change in enthalpy is given by

$\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol$

Entropy is given by

$\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K$

Latent heat of fusion is given by

$L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol$

The latent heat of fusion is 481.76 J/mol

Melting point is given by

$T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K$

Melting occurs at 133.33 K

3 0

3 years ago

1. What is the difference between longitudinal and transverse waves? Compare and contrast

Anvisha [2.4K]

Answer: image to much to type.

Explanation:

8 0

3 years ago

A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.

Zolol [24]

These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is $45^{\circ}$ .

Calling F the force of the wind, and $d=15~km=15000~m$ the distance covered by the boat, the work done by the wind is:
$W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J$

The total time of the motion is $t=1~h=3600~s$ and therefore the power of the wind is
$P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W$

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
$d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m$

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
$MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55$

Finally, the efficiency $\epsilon$ of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
$\epsilon = \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81$

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
$P= \frac{W}{t}$
But we can rewrite the work as
$W=Fdcos\theta$
where F is the force applied, d the displacement of rock and $\theta=60^{\circ]$ is the angle between the direction of the force and the displacement (3 m).
Therefore we can rewrite the power as
$P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta$
where $v=d/t=5~m/s$ is the velocity, Using the data of the exercise, we can then find the force, F:
$F= \frac{P}{v cos\theta} = \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N$

and now we can also calculate the work, which is
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

3 0

3 years ago