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3 years ago

The pic says everything, thank you !

Physics

1 answer:

Alenkasestr [34]3 years ago

4 0

Answer:

1/5 or .2

Explanation:

you put miles over seconds 3/15 then simplify to 1/5 to get miles per second

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Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe

spin [16.1K]

Average velocity =

(displacement) / (time for the displacement)
and
(direction of the displacement) .

Displacement =

(distance from the start-point to the end-point)
and
    (direction from the start-point to the end-point) .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

   300 meters in the direction
   from his house to the neighbor .

His average velocity is

   (300/910) = 0.33 meters per second, in the
   direction from his house to the neighbor .

7 0

3 years ago

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An object of mass 25kg is falling from the height h=10 m. calculate

r-ruslan [8.4K]

Answer:

a=2500J,b=1000K,c=1000J,d=14.142m/s

Explanation:

V²=U²+2gh

V²=0 + 2×10×10=200m/s

a).kinetic energy=(1/2)mv²=(1/2)25×200=2500

potential energy=mgh

p.e=25×10×10=2500J

pe+ke=2500+2500=5KJ

b).mgh=25×10×4=1000J

c). V²=U²+2gh

V²=0+2×10×4

V²=80

kinetic energy=(1/2)mv²

=(1/2)25×80

=1KJ

d). From my first paragraph V²=200

V=√200

V=14.142m/s

6 0

3 years ago

A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?

katrin [286]

Answer:

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

$\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}$

If the period is known, we can find the value of the constant by solving for k:

$\displaystyle k=m\left(\frac{2\pi}{T}\right)^2$

Substituting the given values m=5 Kg and T=2.8 seconds:

$\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2$

k = 25.18 N/m

5 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0

3 years ago

PLEASE HELP FOR BRAINLIEST ANSWER!

Alex787 [66]

I believe the answer is the fourth one, hope this helps

3 0

3 years ago

Read 2 more answers