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Assoli18
5 days ago
5

A center lane with solid and broken yellow lines that is used by vehicles making left turns in both directions is called a:

Physics
1 answer:
Sav [3K]5 days ago
7 0
A left turn lane is designated for vehicles making left turns. Vehicles must completely enter this lane and wait for a clear path before proceeding with the turn. It is not permissible for a vehicle to be partially in and out of the lane, as this obstructs traffic and creates hazardous conditions.
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Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom
ValentinkaMS [3368]

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

D. Value of the centripetal force

3 0
1 month ago
a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w
Keith_Richards [3153]

Answer:

The resulting velocity for him will be 0.187 m/s in reverse direction.

Explanation:

Given:

The mass of the man is, M=75\ kg

The mass of the ball is, m=4\ kg

The initial velocity of the man is, u_m=0\ m/s(rest)

The initial velocity of the ball is, u_b=0\ m/s(rest)

The final velocity of the ball is, v_b=3.50\ m/s

The final velocity of the man is, v_m=?\ m/s

To determine this scenario, we employ the principle of momentum conservation.

This principle states that the total initial momentum equals the total final momentum.

Momentum is calculated by multiplying mass by velocity.

Initial momentum = Initial momentum of the man and the ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of the man and the ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Hence, the total initial momentum equals the total final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign indicates that the man moves backward.

Thus, his final velocity ends up being 0.187 m/s backward.

3 0
19 days ago
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
Yuliya22 [3234]

Answer:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

Explanation:

In this scenario, we determine the initial velocity as follows:

v_i = 7 \frac{m}{s}

The final velocity in this instance can be expressed as:

v_f = 13 \frac{m}{s}

It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:

v_f = v_i +at

Solved for acceleration, we find:

a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}

For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:

v_f = v_ i +a t

Substituting into the equation yields:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

5 0
1 month ago
Ability of the muscles to function effectively and efficiently without undue fatigue
Keith_Richards [3153]

Response:

Physical well-being

Clarification:

7 0
1 month ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Softa [2959]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

Thus,

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now, by applying integration to the equation above

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
1 month ago
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