Which of the following is a balanced equation?

Jerry is conducting an experiment to test friction. His setup is shown below. If he wants to test the friction between different

professor190 [17]

Answer:

The correct option is D

Explanation:

This question is incomplete because of the absence of the setup which as been attached below. The setup shows/determines/tests the friction of wood (which is a block material), since Jerry wants to test the friction between different types of materials, he will have to replace the wooden block with another type of block material of choice so as to determine the friction of that also.

In order to have a comprehensive experiment, Jerry can use 4-5 different types of block material in the course of the experiment.

3 0

3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

Now the horizontal component of velocity:

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

The vertical component of the velocity:

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

Time taken by the projectile to travel the distance of 30 m:

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

Vertical position of the projectile at this time:

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

Now this height should be the maximum height of the tossed apple where its velocity becomes zero.

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

Time taken to reach this height:

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

We observe that $t>t'$ hence the time after the launch of the projectile after which the apple should be tossed is:

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

A heavy piece of hanging sculpture is suspended by a 90 cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at it

kupik [55]

Answer: The mass of the sculpture is 11.8kg

Explanation:

Using the equation of fundamental frequency of a taut string.

f = (1/2L)*√(T/μ) .... (Eqn1)

Where

f= frequency in Hertz =80Hz

T = Tension in the string = Mg

M represent the mass of the substance (sculpture) =?

g= 9.8m/s^2

L= Length of the string=90cm=0.9m

μ= mass density = mass of string /Length of string

mass of string =5g=0.005kg

L=0.9m

μ=0.005/0.9 = 0.0056kg/m

Using (Eqn1)

80= 1/(2*0.9) √(T/0.0056)

144= √(T/0.0056)

Square both sides

20736= T/0.0056

T= 116.12N

Recall that T =Mg

116.12= M * 9.8

M=116.12/9.8

M= 11.8kg

Therefore the mass of the sculpture is 11.8kg

4 0

3 years ago

Two long, parallel wires carry currents of different magnitudes. if the amount of current in each wire is doubled, what happens

Mrac [35]

The force per unit of length between two wires carrying current is
$\frac{F}{L}= \frac{\mu_0 I_1 I_2}{2 \pi r}$
where I1 and I2 are the currents in the two wires, while r is the distance between them.

We can see from the formula that the force is proportional to the product between I1 and I2: $F \sim I_1 I_2$
so, if we double both I1 and I2, we get a factor 4:
$F' \sim (2I_1 )(2I_2)=4 I_1 I_2 =4 F$
so, the force between the wires will be 4 times the original value.

4 0

3 years ago

How many micrograms make 1 kg

Ket [755]

1kg = 1000000000 micrograms

I hope it helps you ;)

4 0

3 years ago