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3 years ago

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In consideration of the above diagram, a projectile is launched at 45 with an initial horizontal velocity (V0x) of 3 m/s. Calcul

ate the Vox which is the actual launch velocity at 45 in m/s.

1 answer:

kumpel [21]3 years ago

8 0

CFISD too huh? :D

also tryna get the answer

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A displacement vector is 34.0 m in length and is directed 60.0° east of north. What are the components of this vector? Northward

PolarNik [594]

Answer:

A) 29.4 m 17.0 m; B) 2 m

Explanation:

If a vector is 34.0 m in length and is directed 60.0° east of north (which means 30.0° over the horizontal), then its coordinates will be:

Horizontal: (34.0 m)cos(30.0°)=29.4 m

Vertical: (34.0 m)sin(30.0°)=17 m

If one person walks 8.0 meters in a straight line and then walks 5.0 meters in another straight line, then the minimum displacement would be to go back over his tracks, displacing himself 8m-5m=3m, while the maximum displacement would be going straight ahead, displacing himself 8m+5m=13m. Any answer outside this interval is impossible (2m).

8 0

3 years ago

A box slides down a frictionless plane inclined at an angle θ above the horizontal. The gravitational force on the box is direct

tamaranim1 [39]

Answer:

D) Vertically.

Explanation:

A free body diagram is used to represent all the forces acting in a body. forces like, the force of gravity as a result of the gravitational interaction between the object and the Earth (W), the frictional force opposite to the movement of the object ( $F_{r}$ ), the normal force due to the plane and the object (N) and the force applied to start the movement in a particular direction (F).

As is show in the free body diagram of the system, W, which is the weight of the body as a consequence of the gravitational force, is at an angle $\theta$ below the inclined plane. that angle between the plane and the x axis is the same that the one of the inclined plane with respect to the horizontal, Since its sides are perpendicular.

Notice how W goes always in the direction to the center of mass of Earth in a vertical path (For comparison see figure (a) and (b)).

4 0

3 years ago

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s

skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0

4 years ago

Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu

Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

$F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N$

$T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N$

$T_x = T*sin(50) = 0.0234 N$

The electric force is given by the expression:

$F = k*\frac{q_1*q_2}{r^2}$

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

$r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m$

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

$F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}$

$q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C$

O 0.247 μC

8 0

3 years ago

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0

3 years ago