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Vikentia [17]
3 years ago
15

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy(KE) when it jumps PLEASE HELP ME

Physics
1 answer:
Kay [80]3 years ago
8 0

Answer:

0.91 J

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the lemming in this problem, when he jumps, we have:

m = 0.0780 kg is the mass

v = 4.84 m/s is the speed

Substituting into the equation, we find:

K=\frac{1}{2}(0.0780)(4.84)^2=0.91 J

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c. Throw the items away from the spaceship.

Explanation:

By the Principle of action and reaction yu can get back to your spacecraft throwing the items away from the spaceship.

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If the mass of a material is 115 grams and the volume of the material is 16 cm3, what would the density of the material be?
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3 years ago
(1 point) A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m
sasho [114]

Answer:

The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)

Explanation:

we have to:

m = mass = 4 kg

k = spring constant = 577 N/m

c = damping constant = 4 N*s/m

The differential equation of motion is equal to:

m(d^2x/dt^2) + c(dx/dt) + k*x = 0

Replacing values:

4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0

Thus, we have:

4*x^2 + 4*x + 577 = 0

we will use the quadratic equation to solve the expression:

x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2)  ± 12i

The solution is equal to:

x = e^(1/2) * (c1*sin12t + c2*cos12t)

x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)

We have the follow:

x(0) = 5

e^0(0*c1 + c2) = 5

c2 = 5

x´(0) = 7

(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7

(-1/2)*(5) + 12*c1 = 7

Clearing c1:

c1 = 0.792

The function is equal to:

x = e^(-t/2) * (0.792*sin12t + 5cos12t)

8 0
3 years ago
Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin
tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a.D=A+B+C

D=3i-3j+i-4j-2i+5j

D=2i-2j

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude=\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2 units

By using the formula \mid r\mid=\sqrt{x^2+y^2}

Direction:\theta=tan^{-1}\frac{y}{x}

By using the formula

Direction of D:\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}

b.E=-A-B+C

E=-3i+3j-i+4j-2i+5j

E=-6i+12j

\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4units

Direction of E=\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}

4 0
4 years ago
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