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4 years ago

How to find the gradient of a velocity time graph

Physics

1 answer:

Sidana [21]4 years ago

7 0

The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

Area of light-blue triangle -
The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: area of triangle = 1⁄2 × base × height so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m. Area under the whole graph The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.This is the total area under the distance-time graph. This area represents the distance covered.

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Which statement correctly describes the difference between an atom and a molecule

Alisiya [41]

Answer:

In a molecule, atoms are bonded together by single, double, or triple bonds. An atom has a nucleus surrounded by electrons.

Explanation:

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3 years ago

A 1000 Kg car approaches an intersection traveling north at 30 m/s . A 1250 Kg car approaches the same intersection traveling ea

Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, $m_1=1000\ kg$

Initial velocity of first car is, $u_1=30\vec{j}\ m/s$

Mass of second car is, $m_2=1250\ kg$

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

$m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s$

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

$m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s$

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, $u_{2y}=0$ .

The magnitude of final velocity is given as:

$|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s$

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ$

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

7 0

3 years ago

I need help with question 6

Irina-Kira [14]

Answer:

She will have to ride her scooter approximately 10 minutes.

8 0

3 years ago

Read 2 more answers

A ball is kicked on a planet the size of Earth but with twice the acceleration due to gravity. If the maximum height of the ball

USPshnik [31]

Answer:

The value $t = 1.995 \ s$

Explanation:

From the question we are told that

The acceleration due to gravity is a = 2g

The maximum height is h= 39 m

Generally from the kinematic equation

$s = ut + \frac{1}{2} at^2$

Here u is the velocity of the ball at maximum height before it start falling and the value is 0 m/s

$39 = 0* t + \frac{1}{2} (2g)t^2$

$39 = (9.8)t^2$

$t = \sqrt{3.9795918}$

$t = 1.995 \ s$

3 0

4 years ago

Two of the types of infrared light, ir-c and ir-a, are both components of sunlight. their wavelengths range from 3000 to 1,000,0

NikAS [45]

The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J

For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J

6 0

3 years ago

Read 2 more answers