Which of these is the most effective way for Leanna to cool down after an intense bike ride

A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil

adelina 88 [10]

<span>14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>

6 0

2 years ago

Read 2 more answers

You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By

Fittoniya [83]

Answer:

The frequency is $f = 0.221 \ Hz$

Explanation:

From the question we are told that

The time taken for it to decay to half its original size is $t = 3.40 \ ms = 3.40 *10^{-3} \ s$

Let the voltage of the capacitor when it is fully charged be $V_o$

Then the voltage of the capacitor at time t is said to be $V = \frac{V_o}{2}$

Now this voltage can be mathematical represented as

$V = V_o * e ^{-\frac{t}{RC} }$

Where RC is the time constant

substituting values

$\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }$

$0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }$

$- \frac{0.5}{RC} = ln (0.5)$

$-\frac{0.5}{RC} = -0.6931$

$RC = 0.721$

Generally the cross-over frequency for a low pass filter is mathematically represented as

$f = \frac{1}{2 \pi * RC }$

substituting values

$f = \frac{1}{2* 3.142 * 0.72 }$

$f = 0.221 \ Hz$

7 0

2 years ago

Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d

kenny6666 [7]

Answer:

428.59 N

Explanation:

Buoyant force, $B=Vg\rho$ where V is volume, g is gravitational constant and \rho is density

$B+F_{upward}=mg$ where $F_{upward}$ is upward force

$Vg\rho_{w}+F_{upward}=mg$

$F_{upward}=mg- Vg\rho_{w}$

$F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}}$ where $\rho_{hippo}$ is the density of hippo

$F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})$

Using g as 9.81

$F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N$

Therefore, the upward force=428.59 N

3 0

2 years ago

Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0

east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero

distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m

3 0

2 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

aleksandrvk [35]

Answer:

Part a)

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

$t = \frac{t_1}{5}$

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

$\omega_f = \omega_i + \alpha t$

$\omega_1 = 0 + \alpha t_1$

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

angular displacement is given as

$\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2$

$\theta = 0 + \frac{1}{2}\alpha t_1^2$

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

As we know that the angular deceleration produced by the brakes is given as

$\alpha_d = - 5\alpha$

now we have

$\omega_f = \omega _i + \alpha t$

$0 = \omega_1 - 5 \alpha_1 t$

$t = \frac{\omega_1}{5 \alpha}$

As we know that

$t_1 = \frac{\omega_1}{\alpha}$

so we have

$t = \frac{t_1}{5}$

6 0

2 years ago