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erastovalidia [21]
3 years ago
7

If a ball has 300 Joules of potential energy at its highest point, how much kinetic energy will it have at its highest speed?

Physics
1 answer:
marshall27 [118]3 years ago
5 0
Pállalas PAOKndndAaaaaaa
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The answer is going to be B
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Equation: -10 40/4^2

Explanation: I multiplied the -10, but idk if that' what you do cause I've never done this. So if you do do that it's -1000, but that is probably wrong. Sorry, I tried.

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Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t
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Read 2 more answers
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
KIM [24]

Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

       t_1 =  250 \ s

The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

7 0
4 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
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