According to the Holland Codes, which career is best suited to someone who scored highest in the investigative category?

Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo

g100num [7]

Answer:the answer is C

Explanation:

3 0

3 years ago

Which of the following are Electromagnetic Waves?

earnstyle [38]

Answer:

b

Explanation:

8 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

A woman walked 115 m. As she did so, her speed increased from 4.20 m/s to 5.00 m/s. How long did it take her to walk this distan

ASHA 777 [7]

Answer:

25 seconds

Explanation:

Assuming the woman is accelerating at a constant rate of $a \;\;m/s^2$ from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.

Let she takes t seconds to cover the distance, s=115 m.

As acceleration, $a=\frac{v-u}{t}=\frac{5-4.2}{t}$

$\Rightarrow at=0.8\cdots(i)$

Now, from the equation of motion

$s=ut+\frac 12 at^2$

$\Rightarrow s=ut+\frac 12 at(t)$

$\Rightarrow 115=4.2t+\frac 12 \times 0.8 t$ [ from equation (i)]

$\Rightarrow 115=(4.2+0.4)t$

$\Rightarrow t= 115/4.6 = 25$ seconds.

Hence, she takes 25 seconds to walk the distance.

5 0

3 years ago

An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is

natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

$C=\frac{K\epsilon _{0}A}{d}$

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

$E=\frac{CV^{2}}{2}$

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0

3 years ago