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azamat
3 years ago
12

Rjejdndjoskdnrksjdjsjsne

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0
Rjejdndjoskdnrksjdjsjsne
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48.3 mph
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3 years ago
A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the
Valentin [98]
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m

Therefore, the displacement would be 0.32 m.
7 0
3 years ago
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a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later
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4 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
3 years ago
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