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Pie
3 years ago
13

A gymnast jumps directly up onto a beam that is 1.5 m high. From the beam, she jumps 0.5 m straight up and lands back on the bea

m in the same spot. What is the total distance she traveled?
A. 0.5 m
B. 1.0 m
C. 1.5 m
D. 2.5 m
Physics
1 answer:
tatiyna3 years ago
5 0

Answer:

D. 2.5 m

Explanation:

1.5 m up to the beam + 0.5 m up + 0.5 m down = 2.5 m total

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The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling \alpha the angle between the vector and the horizontal direction (x), the two sides are related to \alpha by
\tan \alpha  =  \frac{v_y}{v_x}
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3 years ago
How much kinetic energy is in a punch thrown at 30 m/s? The fist and arm weighs 6 lbs. (1 lb= 2.2 kg)
zvonat [6]

Answer:

5940J

Explanation:

KE = 1/2 mv²

KE = 1/2 13.2 * 30²

KE = 6.6*900

KE = 5940J

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3 years ago
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3 0
2 years ago
Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De
My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0
3 years ago
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