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REY [17]
3 years ago
6

A student is leaving the lesson and walking to Maths at a speed of 1m/s. He notices that he is late for the lesson and speeds up

to 3 m/s. It takes him 4 seconds to speed up. Calculate his acceleration.
Physics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

Acceleration = 0.5 m/s²

Explanation:

Given the following data;

Initial velocity, u = 1m/s

Final velocity, v = 3m/s

Time, t = 4 seconds

To find acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Substituting into the equation, we have;

Acceleration = (3 - 1)/4

Acceleration = 2/4

Acceleration = 0.5 m/s²

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Answer:

Q_T=63313.5\ J

Explanation:

Given:

  • temperature of skin, T_s=34^{\circ}C
  • initial temperature of steam vapour, T_v=100^{\circ}C
  • latent heat of steam, L=2256\ J.g^{-1}
  • mass of steam, m=25\ g
  • specific heat of water, c=4190\ J.kg^{-1}.K^{-1}=4.19\ J.g^{-1}.K^{-1}
  • final temperature, T_f=34^{\circ}C

<em>Assuming that no heat is lost in the surrounding.</em>

<u>We know:</u>

Q=m.c.\Delta T

<u>Now the total heat given by the steam to form water at the given conditions:</u>

Q_T=Q_{Lv}+Q_w ..............................(1)

where:

Q_{Lv}= latent heat given out by vapour to form water of 100°C

Q_w= heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

Q_T=m(L+c.\Delta T)

Q_T=25(2256+4.19\times 66)

Q_T=63313.5\ J

is the heat transferred to the skin.

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A quarter is tossed up from the roof of a skyscraper and hits the sidewalk below. Which of the following graphs best shows the v
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3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

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