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3 years ago

Anyone help me do this am giving the brainliest

Physics

2 answers:

bixtya [17]3 years ago

5 0

Answer:

112 km

Explanation:

brainliest plzzzzz i've never got one

lozanna [386]3 years ago

4 0

Explanation:

The first step is to write the 1st condition of equilibrium, which states that the net force acting on the beam is zero:

$F_{net} = 5\:\text{kN} + N_L + N_R - 40\:\text{kN} - 60\:\text{kN} = 0$

where $N_L$ and $N_R$ are reaction forces at the beam supports on the left and right, respectively.

Next, the 2nd condition of equilibrium states that the net torque about a pivot point is zero. Let's choose the location of the left beam support as our pivot point, which will make the torque due to $N_L$ is equal to zero. Let us also assume that the counterclockwise direction produces positive torque. So we can write

$\tau_{net} = (40\:\text{kN})(2\:\text{m}) + (5\:\text{kN})(5\:\text{m}) \\+ N_R(7\:\text{m}) - (60\:\text{kN})(8\:\text{m}) = 0$

$\Rightarrow 80\:\text{kN-m} + 25\:\text{kN-m} + N_R(7\:\text{m}) \\= 480\:\text{kN-m}$

Solving for $N_R,$ we see that

$N_R = 53.6\:\text{kN}$

Putting this value into our 1st equation, we find that the reaction force $N_L$ is

$N_L = 100\:\text{kN} - 5\:\text{kN} - N_R$

$\:\:\:\:\:\:\:\:= 41.4\:\text{kN}$

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3 years ago

I need the answer please help

timurjin [86]

Answer:

A) Inertia

Explanation:

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3 years ago

Read 2 more answers

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using