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3 years ago

If you are traveling around a curve at a constant speed will your velocity change?

1 answer:

4 0

Velocity means [ (speed) and (direction) ].

If you're traveling around a curve, then your direction is
always changing. So your velocity is always changing,
even if your speed isn't.

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Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

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

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

3 years ago

4. If a book falls off the kitchen counter and hits the floor 0.51 seconds later,

Zigmanuir [339]

Answer:

1.2 miles per second

Explanation:

6 0

3 years ago

A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

$F=157\times 0.19\\\\F=29.83\ N$

(B) The force is simply given by :

F = ma

a is acceleration at that instant

$a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2$

6 0

3 years ago

a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how

schepotkina [342]

In order to balance the stick on the pivot, the total "moments" must be equal on both sides. A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction. We have a 25N weight. Where should we hang it ?

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

<em>Distance from the pivot = 3 cm </em>

8 0

3 years ago

Two objects gravitationally attract with a force of 100 N. If the mass of one object is doubled and the mass of the other object

barxatty [35]

Hello!