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geniusboy [140]
2 years ago
10

A book is moved once around the edge of a tabletop with dimensions 1.75 m à 2.25 m. If the book ends up at its initial position,

what is its displacement? If it completes its motion in 23 s, what is its average velocity? What is its average speed?
Physics
1 answer:
uysha [10]2 years ago
6 0

Explanation:

The dimension of the book is 1.75 m × 2.25 m. If the book ends up at its initial position. The displacement of the book is equal to zero as the object reaches to its initial position.

If it completes its motion in 23 s, t = 23 s

Total displacement of the book is equal to its perimeter. It is given by :

d=2(1.75+2.25)=8\ m

The net displacement divided by total time taken is called the average velocity of an object. Here, the displacement is 0. So, average velocity is 0.

The average speed of an object is given by :

v=\dfrac{d}{t}

v=\dfrac{8\ m}{23\ s}

v = 0.347 m/s

So, the average speed of the book is 0.347 m/s. Hence, this is the required solution.

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A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come
Yuliya22 [10]

Answer:

(a)    vo = 24.98m/s

(b)    t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

y=y_o+v_ot-\frac{1}{2}gt^2      (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

v_o=\frac{y-y_o+1/2gt^2}{t}        

v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s

The time that the ball takes to arrive to the ground is 5.09s

5 0
1 year ago
All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is
andre [41]
The acceleration is defined as the rate of change of velocity.
So, if the acceleration is zero, this means that the rate of change of velocity is zero, which also means that the body is moving with constant velocity.
Since we are given that the net forces acting on the body is zero, this means that the body is at equilibrium

Based on this:
<span>All forces on the bullets cancel so that the net force on a bullet is zero, which means the bullet has zero acceleration and is in a state known as equilibrium.

Note that if this constant velocity is equal to zero, then the body would be at rest (not moving)</span>
7 0
1 year ago
Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing
melomori [17]

Answer: Anthony will be warmer after the game.

Explanation :

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing a white shirt. Anthony will be warmer after the game. The black color is a good absorber of radiation and a bad reflector.

The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.

7 0
2 years ago
Read 2 more answers
The ammeter displays a reading of 0.10 A. Calculate the potential difference across the 45 Ω resistor.
NeX [460]

Answer:

V = 45× 0.10= 4.5 volt

...........

5 0
1 year ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

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2 years ago
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