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slavikrds [6]
3 years ago
5

a toboggan loaded with vacationing students (total weight 1300 N) slides down a slope at 30 degrees and there is no friction. wh

at is the acceleration of the toboggan?
Physics
1 answer:
sasho [114]3 years ago
4 0

Answer:

The acceleration of the sliding toboggan is, a = 4.9 m/s²

Explanation:

Given data,

The total weight of the toboggan, W = 1300 N

The slope is, Ф = 30°

The acceleration of a body under the influence of the gravitational field does not depend on its mass, size and shape in the absence of the air resistance.

Therefore,

The acceleration of the toboggan is given by the formula,

                           a = g Sin Ф

Substituting the given values in the above equation,

                           a = 9.8 x Sin 30°

                              = 4.9 m/s²

Hence, the acceleration of the sliding toboggan is, a = 4.9 m/s²

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A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The
Liula [17]
What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be \frac{F}{2} now:

a=\frac{F}{2m}

So the accelaration will also he halved (it's the original acceleratation divided by 2)


 b) The object's mass is halved?
a=\frac{F}{m/2}=a=\frac{F2}{m}

which is the original acceleration times two!! so it will double


c) The force and the object's mass are both halved?
now we have

a=\frac{F/2}{m/2}=a=\frac{2F}{2m}=a=\frac{F}{m}

so they will cancel each other out and the acceleration will stay the same!











5 0
3 years ago
What the unit of work?​
ra1l [238]

Answer:

yes

Explanation:

5 0
3 years ago
A regulation basketball has a 32 cm diameter
Rudik [331]

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

3 0
3 years ago
1. What happens to particle spacing when temperature increase, What we call this process?
enot [183]

Answer:

Particle spacing increases and it's called evaporating

5 0
3 years ago
A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th
finlep [7]

Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

\omega _{0}=\frac{1}{\sqrt{LC}}

\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = \sqrt{R^{2}+\left ( XL - Xc \right )^{2}}

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

7 0
3 years ago
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