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2 years ago

A hamster in it's ball starts at rest and accelerates to 3ms1 in 6 seconds.

Physics

1 answer:

taurus [48]2 years ago

7 0

Answer:9m

Explanation:

Ball starts from rest . Time taken = 6 seconds. Distance travelled by ball. ∴Distance travelled = 9 m

Hope it helps you

Good luck

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The charge that passes a cross-sectional area A=10-4 m2 varies with time according

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Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

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2 years ago

Which one of the following scenarios accurately describes a condition in which resonance can occur? A. A column of air has a hei

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The correct option is B.
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2 years ago

Which of the following describes an air mass with the symbol c t?

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2 years ago

If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt

Alik [6]

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

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2 years ago

Read 2 more answers

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

7 0

3 years ago