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3 years ago

What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.

Physics

1 answer:

Softa [21]3 years ago

7 0

<em>weight = (mass) x (gravity)</em>

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

<em>Weight = 49.05 Newton</em>

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

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

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.

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:

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

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:

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

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3 years ago

PLS ANSWER FAST WILL GIVE BRAINLY TIMED TEST

solong [7]

A=F/m
a=(3000000)/(20000)
a=15 m/s^2

4 0

2 years ago

. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat

AfilCa [17]

Answer:

a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

sin 30 = Wx / W

cos 30 = Wy / W

Wx = W sin30

Wy = W cos 30

Let's write the equations on each axis

X axis

Wx = ma

Y Axis

N- Wy = 0

N = Wy = mg cos 30

N = 2.0 9.8 cos 30

N = 16.97 N

We calculate the acceleration

a = Wx / m

a = mg sin 30 / m

a = g sin 30

a =9.8 sin 30

a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

F -Wx = 0

F = Wx

F = m g sin 30

F = 2.0 9.8 sin 30

F = 9.8 N

5 0

3 years ago

The_____ form acidic compounds with hydrogen.

dsp73

Chlorine forms hydrochloric acid when reacted with hydrogen

7 0

3 years ago

Read 2 more answers

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the mag

nadya68 [22]

Answer:

F = 147,78*10⁻⁹ [N]

Explanation:

By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is

tan β = 0,5/0,7

tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰

cos β = 0,81

d = √ (0,5)² + (0,7)² d1stance between charges

d = √0,25 + 0,49

d = √0,74 m

d = 0,86 m

Now Foce between two charges is:

F = K* q₁*q₂/ d² (1)

Where K = 9*10⁹ N*m²/C²

q₁ = 2,5* 10⁻⁹C

q₂ = 3,0*10⁻⁹C

d² = 0,74 m²

Plugging these values in (1)

F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²

F = 91,21 * 10⁻⁹ [N]

And Fx = F*cos β

Fx = 91,21 * 10⁻⁹ *0,81

Fx =73,89*10⁻⁹ [N]

Then total force acting on charge located at x = 0,7 m is:

F = 2* Fx

F = 2*73,89*10⁻⁹ [N]

F = 147,78*10⁻⁹ [N]

8 0

3 years ago