Violet light has a frequency of 7.26 × 1014 Hz and travels at a speed of 3.00 × 108 m/s.

agasfer [191]

Answer:

For a velocity versus time graph how do you know what the velocity is at a certain time?

Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.

_____________________________________________________

How do you know the acceleration at a certain time?

$Ans: We\ know\ that\ acceleration = \frac{Final\ velocity-Initial\ Velocity}{Time\ taken}$

Hence,

By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.

_________________________________________________________

How do you know the Displacement at a certain time?

Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.

_________________________________________________________

4 0

2 years ago

Geology belongs to which main category of science?

sergey [27]

Geology belongs to the category of Natural Sciences. Natural sciences is the study of natural phenomena. Among the natural sciences, we have physical science, which is the study of non-living systems. Further along this branch, we have earth sciences where Geology belongs.

7 0

3 years ago

Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong):

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

Assuming b = 9.3i + 9.5j (b = 931 + 9.5 is wrong) we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) Asumming (a + b)·b instead a+b·b we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

5 0

3 years ago

One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what

Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

$F_y=ma_y$ (2)

$F_x=ma_x$ (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

$F_y=W+n=0$

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

$F_y=(755)(-9.81)+n=0$

$n=7406.55 N$ (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

$F_x=F+f=ma_x$

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

$1988+f=m(1.36)$

$f=(755)(1.36)-1988=-961.2 N$

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

$|f|=\mu_k n$

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

$\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13$

4 0

3 years ago

When a cup is placed on a table, which force prevents the cup from falling to the ground?

zhuklara [117]

Answer:

B. normal force

Explanation:

Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.

6 0

3 years ago