All categories

3 years ago

_____________ circular motion occurs when an object is traveling with constant speed in a circle.

Physics

1 answer:

emmasim [6.3K]3 years ago

7 0

Your answer is ''Uniform''.

Hope this helps :)

You might be interested in

4. Compare the disturbance before and after the waves pass through the slit.

nataly862011 [7]

Jesus, jesus is always the answer

3 0

4 years ago

Read 2 more answers

What is pitch related to?

Jlenok [28]

Pitch is related to frequency

7 0

3 years ago

Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of

GalinKa [24]

Answer:

D) equal to the flux of electric field through the Gaussian surface B.

Explanation:

Flux through S(A) = Flux through S (B ) = Charge inside/ ∈₀

4 0

3 years ago

A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat

kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0

3 years ago

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

3 years ago