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inysia [295]
3 years ago
6

Does A= 6.4 B=12 C=12.2 Form a RIGHT TRIANLGE?

Physics
1 answer:
Lina20 [59]3 years ago
3 0
To answer this question, you must remember the equation:

a²+b²= c²

(6.4)² + (12)²=   (12.2)²

<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84

This cannot be a triangle

hope this helps
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A 0.141 kg pinewood derby car is moving 1.33 m/s . What is its momentum?
kompoz [17]

momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)

7 0
3 years ago
Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge
xxMikexx [17]

Answer:

new force is 6 times of the initial force.

Explanation:

Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units

If the charge  of Object 1 is doubled and the charge of object 2 is tripled, it means, q_1'=2q_1 and q_2'=3q_2. New force is given by :

F'=\dfrac{kq_1'q_2'}{r^2}\\\\F'=\dfrac{k(2q_1)(3q_2)}{r^2}\\\\F'=6\dfrac{kq_1q_2}{r^2}\\\\F'=6F

So, the new electrostatic force between objects will become 6 times of the initial force.

5 0
2 years ago
A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
What’s the correct answer
White raven [17]
Looks like you simply substitute the length of the femur
8 0
2 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
3 years ago
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