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ICE Princess25 [194]

3 years ago

10

A boy standing on the bridge kicks a stone into the water below. He kicks the stone with a horizontal velocity of 8.06 m/s. It l

ands in the water 6.78 m away from the bridge. How high is the bridge?

1 answer:

Ronch [10]3 years ago

6 0

Time taken to reach water :

$t = \dfrac{D}{v}\\\\t=\dfrac{6.78}{8.06}\ s\\\\t=0.84\ s$

Now, initial vertical speed , u = 0 m/s.

By equation of motion :

$h = ut +\dfrac{at^2}{2}$

Here, a = g = acceleration due to gravity = 9.8 m/s².

So,

$h = 0(t) +\dfrac{9.8\times 0.84^2}{2}\\\\h=3.46\ m$

Therefore, the height of the bridge is 3.46 m.

Hence, this is the required solution.

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Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.

andriy [413]

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

$F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g$

where;

A = L²

$\rho_{\omega} = \rho$

F = ky.

Then,

$Ay \rho g = ky$

$L^2y \rho g = ky$

Divide both sides by y

K = ρL²g

3 0

3 years ago

A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and

natima [27]

The upward force exerted on the board by the support is 530.8 N.

<h3>Upward force exerted on the board by the support</h3>

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

Learn more about upward force here: brainly.com/question/6080367

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8 0

2 years ago

While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38° with the vertical. While driving

Natali [406]

Answer: $21.33^{\circ}$

Explanation:

Given

velocity of driver $v_1$ =25 m/s w.r.t ground towards north

driver observes that rain is making an angle of $38^{\circ}$ with vertical

While returning $v_2$ =25 m/s w.r.t. ground towards south

suppose $u_1$ =velocity of rain drop relative car while car is going towards north

$u_2$ =velocity of rain drop relative car while car is going towards south

z=vector sum of $u_1 & v_1$

Now from graph

$\tan 38=\frac{v_1+v_2}{u_2}$

$u_2=\frac{25+25}{\tan 38}=64 m/s$

$z=\vec{u_2}+\vec{v_2}$

therefore magnitude of z is given by

$|z|=\sqrt{u_2^2+v_2^2}$

$|z|=\sqrt{64^2+25^2}$

$|z|=68.70 m/s$

$\tan A=\frac{v_2}{u_2}$

$\tan A=\frac{25}{64}=0.3906$

$A=21.33^{\circ}$

Thus rain drops make an angle of $21.33^{\circ}$ w.r.t to ground

6 0

3 years ago

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

Maru [420]

Answer:

$q=7.965*10^-^6C$

Explanation:

From the question we are told that

Altitude of $d_1 390m$

Magnitude $M_1=60.0 N/C$

Altitude of $d_2=240 m$

Magnitude is $M_2= 100 N/C$

Distance of cube $d_c=150 m$

Generally the flux $\phi$ is mathematical given as

$\phi=60(150)^2cos180+100(150)^2*cos0$

$\phi=-9*10^5$

Generally Quantity of charge q is mathematically given as

$q=\varepsilon _0 *\phi$

$q=8.85*10^-^1^2 *9*10^5$

$q=7.965*10^-^6C$

5 0

3 years ago

I understand what makes a kettle so efficient, but what are some improvements you can make to a kettle to increase it’s energy e

shusha [124]

Answer:

Reheat the cold cup of tea or coffee in the microwave. ...

Make iced tea or coffee with what's left. ...

Use a thermos for either tea or coffee.

3 0

3 years ago