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kodGreya [7K]
3 years ago
5

I will mark as the brainliest answerplz 8,9,10​

Physics
1 answer:
blagie [28]3 years ago
5 0

Answer:

8.  acceleration = \dfrac{d(velocity)}{d(time)}  = 1 unit .

9. acceleration = \dfrac{d(velocity)}{d(time)}  = -1 unit.

10. acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units.

Explanation:

8. i) acceleration = velocity / time

  ii) In this figure velocity = time

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)} = 1 unit .

9. i) acceleration = velocity / time

  ii) In this figure 4 = m + 5, therefore m = -1

     therefore velocity = (-0.5 \times time) + 5

  iii) therefore acceleration = \dfrac{d(velocity)}{d(time)}  = -1 units.

10.) velocity is constant at 2

     therefore acceleration = \dfrac{d(velocity)}{d(time)}  = 0 units

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When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea
alekssr [168]

Answer:

The horizontal distance traveled by the projectile is 15.23 m.

Explanation:

Given;

angle of projection, θ = 25⁰

initial velocity of the projectile, u = 15 m/s

time of flight, t = 1.12 s

The the travelling path of the object is calculated as the range of the projectile

R = u_x t\\\\R = (15\ Cos \ 25^0) \times 1.12\\\\R = 13.595 \times 1.12\\\\R = 15.23 \ m

Therefore, the horizontal distance traveled by the projectile is 15.23 m.

8 0
3 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has
Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

<u>r = 0.5 m</u>

8 0
3 years ago
What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?
LenaWriter [7]

Answer:

Hello,~There!

What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

<h2><u>33% According to the Graph</u></h2>

Hope this helps!

6 0
3 years ago
Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the
Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}

The force can be calculated by the formula

F=k\frac{q1q2}{(r^{\prime})^2}

Here, k is the constant whose value is

k=9\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the force will be

\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}

7 0
1 year ago
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