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weeeeeb [17]
3 years ago
7

What is the speed of a bobsled whose distance-time graph indicates that it traveled 124m in 26s

Physics
1 answer:
krek1111 [17]3 years ago
5 0
It's average speed during that 26 seconds was about 4.77 m/s. Without seeing the graph, we can't tell if it was going faster or slower at any particular time during that period. All we can tell is its average for the full interval.
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a mass of 10kg is placed on a horizontal table with coefficient of friction is 0.5. if the mass is static, determine weight of t
zloy xaker [14]

Answer:

10

Explanation:

5 0
3 years ago
A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3
Klio2033 [76]

Answer:

\theta=30^{\circ}

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

F=ILB\ sin\theta

sin\theta=\dfrac{F}{ILB}

sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}

\theta=30^{\circ}

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0
3 years ago
What is the trend noticed with the inner planets
lutik1710 [3]
The inner planets are closer to the Sun and are smaller and rockier. ... The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars.Apr 23, 2014
7 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
Question 10 of 30
Yuliya22 [10]

Answer:

Technology, the application of scientific knowledge to the practical aims of human life or, as it is sometimes phrased, to the change and manipulation of the human environment.

Explanation:

3 0
2 years ago
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