Hi can someone please answer the number 2 question i really need it thank you ❤️

As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a

Annette [7]

Im pretty sure it’s a because it makes more sense you know?.

5 0

2 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

What is cosmic background radiation?

Kobotan [32]

Cosmic background radiation is electromagnetic radiation from the sky with no discernible source. The origin of this radiation depends on the region of the spectrum that is observed.

6 0

3 years ago

Read 2 more answers

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds 

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

7 0

3 years ago

A sphere has a radius of 3.9cm and a density of 7.58 g/cm cubed. What is the mass?

snow_lady [41]

Answer:

m=ρV

V=4/3 * pi * r3

V=1.3 * 3.14 * 3.9^3

V=242.14 cm^3

m=7.58 * 242.14

m=1.8 kG

Explanation:

1. We calculate volume for sphere.

2. Then we calculate mass of sphere.

3 0

3 years ago

Hi can someone please answer the number 2 question i really need it thank you ❤️​

Hi can someone please answer the number 2 question i really need it thank you ❤️