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mart [117]
3 years ago
9

A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)

Physics
1 answer:
MrRissso [65]3 years ago
4 0

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

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A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
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Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

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Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

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7 0
3 years ago
) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)
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To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

\tau = \vec{F} \times \vec{r}

\tau = (8i+6j)\times(-3i+4j)

\tau = (8*4)(i\times j)+(6*-3)(j\times i)

\tau = 32k +18k

\tau = 50 k

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}

cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}

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5 0
3 years ago
The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is
kodGreya [7K]

Answer:

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Explanation:

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where,

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Therefore,

\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\

\frac{T_t}{T_c} = 1.32

8 0
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