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4 years ago

Find the acceleration of a body whose velocity increases from 11ms-1 to 33ms-1 in 10 seconds

2 answers:

8 0

Acceleration is equivalent to the rate of change of velocity.

Calculate the difference of velocity, and divide by the total time it took for the change to happen.

$(33 \frac{m}{s} - 11 \frac{m}{s} ) \div 10seconds \\ = 2 \frac{m}{ {s}^{2} }$
That is the answer.

Alex777 [14]4 years ago

7 0

As we know that
a= dell V/t
then put values
a= 2ms/^2

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What current flows through a 2.60-cm-diameter rod of pure silicon that is 20.5 cm long, when 1.00 ✕ 103 V is applied to it? (Suc

Virty [35]

Answer:

1.13 mA

Explanation:

Length of wire L = 20.5 cm = 0.205m

Radius of wire r = 2.60/2 = 1.3cm = 0.0130m

Voltage V = 1 × 10³ V

Resistivity of pure silicon p = 2300 Ohms • m

Cross sectional area of the wire

A = pi × r² = pi × (0.013)² = 5.307 × 10 ^-4 m²

Resistance of the material

R = p• L/A

= 2300 • 0.205/5.307 × 10^-4 = 0.888 × 10⁶ Ohms

Using Ohms Law

R = V/ I

I = V/R

I = 10³/0.888 × 10⁶

= 0.001126 A

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4 years ago

A space probe that weighs 5,630 N on the Earth has a weight of 4,504 N on another planet. What is the gravitational field streng

Arturiano [62]

Answer:

6,000 gravity force is your answer another question

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2 years ago

Name 8 examples of Convection.

REY [17]

1: Boiling Water
2: Ice Melting
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3 years ago

A point charge Q is placed at the center of a conducting spherical shell (inner radius a, outer radius b). What is the electric

sp2606 [1]

Answer:

a) $E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

b)E=0

c) $E=\dfrac{Q}{\varepsilon _o\times 4\pi r^2}\ N/C$

Explanation:

Given that

A point charge Q is placed at the center of a conducting spherical shell .Due to this - Q charge will induce on the inner sphere surface and +Q will induce on the outer sphere surface .

a) r < a

At a radius r ,from gauss theorem

$E.ds=\dfrac{q_i}{\varepsilon _o}$