Which three types of electromagnetic waves carry most of the Sun's energy that strikes Earth?

How can you determine the rate of a chemical reaction?

DanielleElmas [232]

Take a look at a simple reaction like the one below:

In this reaction some reactant A is turned into some product B. The rate of reaction can be represented by a decrease in concentration of A over time or as the increase of B over time. This is written:

6 0

3 years ago

Why is vehicle systems forensics useful today? Cars are less computerized than before More people own cars Devices like smartpho

kvasek [131]

Answer:

Devices like smartphones can interface with cars and leave evidence

Explanation:

Vehicle system forensic relates to digital data stored in a vehicles system.

Bluetooth connection times can be used to figure out at what time the owner was near his car. e.g. between a smart-watch and car infotainment system

6 0

3 years ago

8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

nasty-shy [4]

Answer: The ball is travelling with a speed of 5.5 m/s after hitting the bottle.

Explanation:

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

$m_1,u_1\text{ and }v_1$ are the mass, initial velocity and final velocity of ball.

$m_2,u_2\text{ and }v_2$ are the mass, initial velocity and final velocity of bottle.

We are given:

$m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s$

Putting values in above equation, we get:

$(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s$

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

5 0

3 years ago

A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc

Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

$v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s$

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ = $\frac{-17.2}{16.6} =-1.036$

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0

3 years ago

A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th

lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

7 0

3 years ago

Read 2 more answers