You have a 6.00 V power supply, and

What is the transfer of energy to an object using a force that causes the object to move in the direction of the force?

anastassius [24]

C. Work
$w = f \times d$
*Work is measured in Joules.

4 0

3 years ago

A stone is tied to a 0.85-meter cord. it is swung in a circle at a constant rate of 6.0 m/s. what is the centripetal acceleratio

Andrei [34K]

Answer:42.4m/s^2

Explanation:

Velocity(v)=6m/s

Radius(r)=0.85 meter

Centripetal acceleration=(v x v) ➗ r

Centripetal acceleration=(6 x 6) ➗ 0.85

Centripetal acceleration=36 ➗ 0.85

Centripetal acceleration=42.4

5 0

3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

$f_r = 150.47 N$

8 0

4 years ago

PLS THIS IS DUE IN 2 MINUTES

Tom [10]

Answer:

The toy car. An object that isn't moving has no momentum

Explanation:

3 0

3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago