Ask question

Ask question

All categories

3 years ago

15

Which of these is an example of acceleration? A a bicyclist turning around a corner B a car traveling south with its cruise cont

rol set C a bus traveling north at 25m/s D a helicopter waiting to take off

1 answer:

Airida [17]3 years ago

4 0

The answer is A.

Acceleration means when you go faster in speed. Since the bicyclist is turning, he has to slow down but when he finishes turning, he is going to pedal harder and gain more speed. This is called Acceleration.

Best of Luck!

You might be interested in

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration

KIM [24]

Answer:

A) B = 0.009185 T

B) Drection is negative y-direction

Explanation:

A) We are given;

Speed(v) = 2.5 x 10^(7) m/s

Acceleration (a) = 2.2 x 10^(13) m/s²

We also know that charge of proton(q) = 1.6 x 10^(-19)

Mass of proton(m) = 1.67 x 10^(-27)

Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;

F = qvBsinθ = ma

Since perpendicular, θ = 90°

And so, sinθ = sin 90 = 1

Thus, qvB = ma

Making B the subject gives;

B = ma/qv

B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))

= 0.009185 T

B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction

4 0

4 years ago

Read 2 more answers

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

True or False: The basketball should be dribbled below the waist.

zimovet [89]

True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.

7 0

2 years ago

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ

LenaWriter [7]

The orbital period of the satellite[T] is given as $2*10^{6} S$ .

The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V] written as -

V $=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

$V=\frac{2*3.14*10^{12} }{2*10^{6} }$

$= 7.85*10^{6} m/s$

There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

⇒ $GM=gR^{2}$ where g is the acceleration due to gravity of that central object.

Hence $V=\sqrt{\frac{gR^{2} }{R} }$

⇒ $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.

7 0

3 years ago

Read 2 more answers

How could you increase the total energy of a system consisting of a jar full of water?

grin007 [14]

Answer:

D.) Heat the jar in a microwave oven.

Explanation:

5 0

3 years ago

Read 2 more answers