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joja [24]
3 years ago
14

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

by it in this time
Physics
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}

\boxed{\sf Distance \ covered \ (s) = 100 \ m}

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

\sf (i) \ From \ 1^{st} \ equation \ of \ motion:

\sf \implies v = u + at

\sf \implies 20 = 0 + a(10)

\sf \implies 10a = 20

\sf \implies \frac{10a}{10}  =  \frac{20}{10}

\sf \implies a = 2 \: m/ {s}^{2}

\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:

\sf \implies s = ut +  \frac{1}{2} a {t}^{2}

\sf \implies s = (0)(10) +  \frac{1}{2}  \times 2 \times  {(10)}^{2}

\sf \implies s =  \frac{1}{ \cancel{2}}  \times  \cancel{2} \times  {(10)}^{2}

\sf \implies s =  {10}^{2}

\sf \implies s = 100 \: m

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allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

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3 years ago
One molecule of bromine (Br2) and two molecules of potassium chloride (KCI) combine in a reaction. How many atoms are in the pro
Liula [17]

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Let's consider the balanced equation for the reaction between 1 molecule of bromine and 2 molecules of potassium chloride. This is a single replacement reaction.

Br₂ + 2 KCl ⇒ 2 KBr + Cl₂

We obtain as products, 2 molecules of potassium bromide and 1 molecule of chlorine.

  • 1 molecule of KBr has 2 atoms, so 2 molecules contribute with 4 atoms.
  • 1 molecule of Cl₂ has 2 atoms.
  • The 4 atoms from KBr and the 2 atoms from Cl₂ make a total of 6 atoms.

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Learn more: brainly.com/question/21850455

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2 years ago
A cylinder with a piston contains 0.250 mol of oxygen at2.40x105 Pa and 355 K. The oxygen may be treated as an ideal gas.The gas
Nikitich [7]

Answer:

= 285 Joules

Explanation:

a) answer can be found out in attachment

(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:

p V = nR T          to see that the temperature also doubles.

.So...   temp for isothermal compression =   355×2 = 710 K

.(c)   The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or

max pressure = 2×240000 Pa =  480000 Pa  =   4.80 x 10^5 Pa

(d) total work done by the piston = workdone during isothermal compression - work done during expansion =

=  nRT ln(V initial / V final)-p (V initial - V final)

=   nRT ln(2) - nR(T final - T initial)  

= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)

=     285 Joules

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