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Ganezh [65]
3 years ago
15

Read the scenario. A horse accelerates +2 mi/h/s south. Which option defines the horse’s acceleration?

Physics
1 answer:
Yanka [14]3 years ago
3 0

Answer:

D) The horse’s velocity increases by 2 mi/h every 1 s as the horse moves south

Explanation:

Let's analyze all options:

A. The horse’s speed changes to 2 mi/h as the horse moves south The horse's speed is not 2 mi/h, but it increases 2 mi/h every second. So, this is false.

B. The horse’s speed decreases by 2 mi/h every 1 s. It would only decrease if the horse was moving north. Otherwise this is false.

C. The horse’s position decreases by 2 mi every 1 s as the horse moves south. It is not horse's position but its velocity what changes 2mi/h every second. Also false.

D. The horse’s velocity increases by 2 mi/h every 1 s as the horse moves south. Since given acceleration is positive, you can say that its velocity increases 2mi/h every second.  So, this is true

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What is the force on a box that is being pushed to the right with 50n of force while, at the same time, being pushed to the left
elena-s [515]
I think the question meant to say net force on the box. Since force is a vector, the direction matters. 20N left is negated completely by the 50N right, which means the net force is 50N-20N to the right, 30N. 
7 0
3 years ago
The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve
Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

\varepsilon _{long}=-0.148

So the axial pressure

P=E\times \varepsilon _{long}

P=5\times 0.148

P=740 KPa

The movement in the sleeve

\Delta =\varepsilon _{long}\times L

\Delta =0.148\times 50

Δ=7.4 mm

6 0
2 years ago
1 example of a conductor and 1 example of a insulator in your EVERYDAY world.
ratelena [41]

Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

4 0
2 years ago
Name two ways to reduce friction and two ways to increase it.
Maurinko [17]
Ways to increase friction 

<span>- increase the roughness of the contact materials </span>
<span>- increase the pressure on the contact </span>


<span>Ways to decrease friction </span>

<span>- float the moving body on air </span>
<span>- suck out any air </span>
4 0
2 years ago
When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s
d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(2)^2 = 0.5 J

2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(4)^2 = 2 J

4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(5)^2 = 3.125 J

5) when speed is 6 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(6)^2 = 4.5 J

3 0
3 years ago
Read 2 more answers
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