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3 years ago

Timmy bikes to school 3 kilometers east and 2 kilometers south. What is Timmy's final displacement?

Physics

1 answer:

vovikov84 [41]3 years ago

8 0

Answer:

3.6km South East

Explanation:

Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:

Pythagoras Theorem:

a^2+b^2=c^2

(2)^2+(3)^2=c^(2)

4+9=c^2

Square root 13 = c

c=3.6km (1dp)

The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).

Hope this helped!

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How many kilocalories (Calories) does the snack bar contain?

laiz [17]

Answer:

See the explanation below.

Explanation:

Attached is a picture of a chocolate bar that can be easily found on the market. In this picture, we see the table of contents with the amount of energy that depends on the mass in grams of the chocolate bar.

We see that for the bar size of 100G, the energy value is 483 [kcal].

We can see that the energy depends on the size of the chocolate bar.

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3 years ago

Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble

Llana [10]

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

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3 0

3 years ago

A 48.5 kg student runs down the sidewalk and jumps with a horizontal speed of 4.25 m/s onto a stationary skateboard. The student

storchak [24]

<u>Answer:</u>

2.39 kg

<u>Explanation:</u>

There is conservation of momentum here in this problem so we will use the following problem:

$m_1u_1+m_2u_2=(m_1+m_2)v$

where the mass of the student $m_1$ is 48.5 kg,

the mass of the skateboard $m_2$ is $m_2$ kg,

the initial speed of the student $u_1$ is 4.25 m/s; and

the speed of the student and skateboard $v$ is 4.05 m/s.

So substituting the given values in the above formula to get:

$(48.5*4.25) + (m_2 * 0) = (48.5 + m_2 ) * 4.05$

$206.125=196.425+4.05m_2$

$206.125 - 196.425 = 4.05m_2$

$m_2=\frac{9.7}{4.05}$

$m_2=2.39$

Therefore, the mass of the skateboard is 2.39 kg.

7 0

3 years ago

assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th

finlep [7]

Answer:

A. xmax = 131.49 m

B. t = 8.74 s

C. ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

$y=y_o+v_osin\theta-\frac{1}{2}gt^2$ (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

$0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2$ (2)

You use the quadratic formula to obtain the value of t:

$t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s$

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

$x_{max}=v_ocos\theta t$

$x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m$

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

$y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}$ (3)

By replacing in the equation (3) the values of all parameters you obtain:

$y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m$

The maximum height reached by the cannon ball is 220.33m

3 0

3 years ago

A 5000 kg truck traveling at 60 m/s stops in 5 seconds. How much friction was between the truck's tires and the ground?

oksano4ka [1.4K]

Answer:

60000N

Explanation:

acceleration is change in velocity

a =(v-u)/t where a is acceleration u is initial velocity and v is final velocity

a = (0-60)/5 = -60/5= - 12m/s^2

here minus sign shows that body is decelerating and force is force of friction Now f = ma here f is force of friction m is mass and a is acceleration

f= 5000×- 12= -60000N

MINUS SIGN HERE SHOWS FORCE OF FRICTION

Hence force of friction is 60000N

8 0

4 years ago