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4 years ago

A cart is given an initial velocity of 5.0 m/s and

Physics

1 answer:

larisa86 [58]4 years ago

5 0

Answer:

cart displacement is 66 m

Explanation:

given data

velocity = 5 m/s

acceleration = 2 m/s²

time = 6 s

to find out

What is the

magnitude of cart displacement

solution

we will apply here equation of motion to find displacement that is

s = ut + 0.5×at² .............1

here s id displacement and u is velocity and a is acceleration and time is t here

put all value in equation 1

s = ut + 0.5×at²

s = 5(6) + 0.5×(2)×6²

s = 66

so cart displacement is 66 m

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How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

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3 years ago

Equal masses of he and ne are placed in a sealed container. What is the partial pressure of he if the total pressure in the cont

EastWind [94]

Answer:

6 atm.

Explanation:

Let the mass of both be m

Then moles of He = m/ 4

Moles of Ne = m/ 20

mole fraction of He = Moles of He/ Total moles = m/4/ (m/4 + m/20) = 0.25 m/0.3m = 0.83

Pressure of He = Mole fraction×total pressure = 0.83 × 6 atm = 5 atm

6 0

3 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

3 years ago

Are basins a collection of smaller watersheds

mihalych1998 [28]

Explanation:

both are areas of land that drain to particular water bodies such as lakes

7 0

4 years ago