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3 years ago

8

A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before

striking the water

1 answer:

DENIUS [597]3 years ago

4 0

Answer:

14.05m/s

Explanation:

We are given that

Horizontal speed of diver= $v_x=1.2m/s$

Distance, y=-10 m

It is taken as negative because the diver goes downward.

There is no air resistance therefore, there is acceleration due to gravity .

We know that

$g=-9.8 m/s^2$

Initial vertical velocity, $u_y=0$

$v^2-u^2=2as$

Using the formula

$v^2_y-0=2\times (-9.8)(-10)$

$v_y=\sqrt{2(-9.8)(-10)}$

$v_y=14 m/s$

$v=\sqrt{v^2_x+v^2_y}$

Using the formula

$v=\sqrt{(1.2)^2+(14)^2}=14.05m/s$

Hence, the speed of diver before just striking the water=14.05m/s

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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P

MArishka [77]

Answer:

$F \approx 19.5 N$

Explanation:

From the question we are told that:

Pressure of $P_{CO_2}=1.50 * 105 Pa.$

Bottle cap area $A_b= 4.40 * 10-4 m^2$

Generally the equation for Resultant pressure $P_r$ is give as is mathematically given by

$P_r=P_{CO_2}-P_a$

Where

$P_a=atmospheric\ pressure = 1.013*10^5 pa$

$P_r=1.50 * 105 Pa-1.013*10^5 pa$

$P_r=0.487*10^5 pa$

Generally the equation for Force exerted by screw F is give as is mathematically given by

$F = P*A\\F = 0.487*10^5*4.00*10^-4\\ F = 19.48 N$

$F \approx 19.5 N$

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2 years ago

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lilavasa [31]

Explanation:

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2 years ago

The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i

attashe74 [19]

Let the vector position of the object in the (x-y) plane be
$\vec{r} = x \hat{i} + y \hat{j}$

The applied force is
$\vec{f} = -12.5 \hat{i}
$

By definition, the applied torque is
$\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}$

Answer: $12.5y \, \hat{k}$

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3 years ago

How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners

Natasha_Volkova [10]

Answer:

$Potential\ Energy=Work \ Done=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V= $\frac{kq}{r}$

So,

$Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}$

Where:

k is Coulomb Constant= $8.99*10^9 Nm^2/C^2$

q is the charge on electron= $-1.6*10^-19 C$

r is the distance= $3.0*10^{-10}m$

For 3 Electrons Potential Energy or work Done is:

$Potential\ Energy=3*\frac{kq^2}{r}$

$Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J$

Work done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is $2.301*10^{-18} Joules$

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3 years ago

morpeh [17]

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

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2 years ago