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natulia [17]
3 years ago
14

Fbejfsjnvfdjbesjkbf dsjlbc dsjlc

Physics
2 answers:
pav-90 [236]3 years ago
7 0

Answer:

???

Explanation:

mojhsa [17]3 years ago
4 0

Answer:

the question, plz?

Explanation:

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When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of
pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0
3 years ago
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

4 0
2 years ago
A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
Red clothes will---the red light.A)Reflect B)refract C)Absorb D)transmission E)Dispersion
posledela

Red clothes look red  because they REFLECT the red light, and absorb light of other colors.

7 0
3 years ago
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