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3 years ago

7

What is the average speed of this car over the entire journey ?

1 answer:

lesya [120]3 years ago

4 0

Answer: 0.83333333333 km/h

Explanation: give brainliest

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Calculate the acceleration of a bus full speed changes from 6 miles to 12 miles over a period of three seconds

Jet001 [13]

Answer:

0.90m/s²

Explanation:

Given parameters:

Initial speed = 6miles/hr

Final speed = 12miles/hr

Time taken = 3 seconds

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time. It is mathematically given as:

Acceleration = $\frac{Final speed - Initial speed}{Time}$

We need to convert miles/hr to meters/seconds

Initial speed = 6 x $\frac{miles }{hour}$ x $\frac{1hr}{3600s}$ x $\frac{1609.34m}{1mile}$

= 2.68m/s

Final speed = 12 x $\frac{1609.34}{3600}$ = 5.37m/s

Acceleration = $\frac{5.37 - 2.68}{3}$ = 0.90m/s²

4 0

3 years ago

DochEvi [55]

Answer:

3.83×10⁻¹⁹ J

Explanation:

The energy of a photon can be found using the formula: E=hc/λ

h is Planck's Constant which is approximately 6.63×10⁻³⁴
c is the speed of light which we'll round to 3.00×10⁸
λ is the wavelength of the photon. You typed 5200 but that's not even within the visible light spectrum's wavelength range, so I'm going to assume you meant 520 nm; or 5.20×10⁻⁷ m.

Plugging all that info in should give approximately 3.83×10⁻¹⁹ Joules of energy.

(You could get slight variations depending on what you chose to round up or down or not at all.)

5 0

3 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

Question 2 (1 point)

Tju [1.3M]

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

7 0

3 years ago

Which of the following activities belongs on the top of the physical activity for pyramid

jasenka [17]

Playing Vedio Games. :)

7 0

4 years ago