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3 years ago

As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

Physics

1 answer:

vlada-n [284]3 years ago

7 0

Answer:

$a=-0.33\ m/s^2$

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

$\displaystyle a=\frac{\Delta v}{t}$

Or, equivalently

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

$\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2$

The negative sing of a indicates there is deceleration or decreasing speed

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A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

6 0

3 years ago

Suppose you hear a clap of thunder 14.4 s after seeing the associated lightning strike. The speed of light in air is 3.00 ✕ 108

Marina86 [1]

Answer

Given,

Time to hear the clap = 14.4 s

speed of the light = 3 x 10⁸ m/s

Speed of sound = 343 m/s

a) distance where lightning strike

D = s x t

D = 14.4 x 343

D = 4939.2 m

b) No, we do not need to know the value of speed of light. Because we need to calculate the distance where we hear the sound. To calculate that we need to know the speed of sound.

5 0

3 years ago

The Himalayas are the highest mountain in the world (positive). grammar question

andrew-mc [135]

Answer:

positive degree:

No other peak of the Himalayas is as high as Mount everest

7 0

3 years ago

A 43 kg box is being pushed and pulled by

m_a_m_a [10]

Answer:

a = - 0.209 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

We will take the positive forces to the right and the negative forces to the left.

$155-277+113=43*a\\-9=43*a\\a = - 0.209[m/s^{2} ]$

The negative sign means that the box accelerates in a negative direction to the left.

4 0

3 years ago

How to Answer questions 12 and 13

zubka84 [21]

12) The weight of the plane is $7.35\cdot 10^5 N$

13) The lift provided by the wing is $7.35\cdot 10^5 N$

Explanation:

12)

The weight of an object is equal to the force of gravity acting on the object. Near the Earth's surface, it can be calculated as

$W=mg$

where

W is the weight

m is the mass of the object

g is the acceleration of gravity

In this problem, we know that the mass of the plane is

m = 75 tonnes

Since 1 ton = 1000 kg, the mass in kg is

m = 75,000 kg

Also, the acceleration of gravity is

$g=9.8 m/s^2$

Therefore, the weight of the plane is:

$W=(75,000)(9.8)=7.35\cdot 10^5 N$

13)

We can solve this question by applying Newton's second law along the vertical direction of motion. In fact, the net force acting along the vertical direction must be equal to the product between the mass of the plane and the vertical acceleration:

$\sum F_y = ma_y$