All categories

3 years ago

Can someone give me some tips of how to study. I have already tried writing everything in my lesson but it doesn't seem to work

1 answer:

3 0

Highlighter, color coding sections that go together, flash cards, relaxing with someone while they read out questions ( this will help with testing anxiety too). If you get distracted easily use games to study or small sayings instead of the usual white and black sheet of paper.

You might be interested in

A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between

Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

$a = \dfrac{F}{m}$

Where, F = frictional force

m = mass

Put the value into the formula

$a=\dfrac{\mu mg}{m}$

$a=\mu g$

$a=0.229\times9.8$

$a=2.244\ m/s^2$

We need to calculate the speed of the sled

Using equation of motion

$v^2=u^2-2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

$v ^2=(5.49)^2-2\times2.244\times3.89$

$v=\sqrt{(5.49)^2-2\times2.244\times3.89}$

$v=3.56\ m/s$

Hence, The speed of the sled is 3.56 m/s.

8 0

3 years ago

Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.91 m, a 2.95-kg object is at +2.43 m, a 2.41-kg

Anna [14]

Answer:

0.958 m

Explanation:

So the total mass of the system is

M = 1.93 + 2.95 + 2.41 + 3.99 = 11.28 kg

let y be the distance from the center of mass to the origin. With the reference to the origin then we have the following equation

$My = m_1y_1 + m_2y_2 +m_3y_3 + m_4y_4$

$11.28y = 1.93*2.91 + 2.95*2.43 + 2.41*0 + 3.99*(-0.496) = 10.806$

$y = \frac{10.806}{11.28} = 0.958 m$

So the center of mass is 0.958 m from the origin

3 0

2 years ago

A ball of mass 0.35 kg hangs straight down on a string of length 13 cm. It is then swung upward, keeping the string taut, until

Anni [7]

Answer:

PE=0.29J

Explanation:

According to the description, there is a angle and in point swung upward of 70°

So,

$Y=13*10^{-2}*cos(70) \\Y=0.0444m$

Appling the equation of Potential Energy we have,

$PE=mgh\\PE=(0.35)(9.8)(0.13-0.0444)\\PE=0.29J$

6 0

3 years ago

What is the weight of a bag that has the mass of 18.0kg

Novay_Z [31]

39.6832 pounds would be your answer

8 0

3 years ago

Read 2 more answers

A boat with an anchor on board floats in a swimming pool that is somewhat wider than the boat. Does the pool water level move up

pentagon [3]

Answer:

a) moves down

b) moves down

c) level remains same

Explanation:

Given that the anchor is initially on the floating boat.

In this condition initially the the volume of water $(V_w_i)$ displaced is to balance its weight.

Now,

$W_a=W_w$

$V_a.\rho_a.g=V_w_i.\rho_w.g$

$\frac{\rho_a}{\rho_w} =\frac{V_w_i}{V_a}$

We've, the density of steel $= 7850\ kg.^{-3}$ and the density of water $= 1000\ kg.^{-3}$

$\therefore V_w_i=7.85\times V_a$

When the anchor is dropped into water:

The volume of water displaced be $(V_w_f)$ which will be equal to the volume of anchor since it is immersed into it.

$V_a=V_w_f$

$\therefore V_w_i>V_w_f$ ...................(1)

So the level of water falls when the anchor is dropped into water.

Now, when the anchor is thrown on the ground the water has now less weight to balance so the water level falls down.

When the cork on the from the boat is dropped into the water and it still floats then it must displace same amount of water, hence there should be no change in the water level.

6 0

3 years ago