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2 years ago

A horse runs 350 meters in 23 seconds. What was the average speed of the horse? Round your answer to a tenth

Physics

1 answer:

OLga [1]2 years ago

5 0

Answer:

Horse/Speed

55 mph

rounded to the tenth? either 60 or 50

but 350 would stay like that i believe!

Explanation:

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At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago

You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the sur

borishaifa [10]

Answer:

D. Half as great

Explanation:

Since we know that the friction force between the surface of crate and ground is given as

$F_f = \mu F_n$

so here we know that

$\mu$ = friction coefficient between two surfaces which depends on the effective contact area between two surfaces

$F_n$ = normal force due to the object

So when we turn the object on another side such that the surface area is half then the friction coefficient will become also half

So here the friction force will also reduce to half

so correct answer will be

D. Half as great

3 0

2 years ago

If a tank contains water 4 m deep, what is the pressure at the bottom of the tank? (Neglect the atmospheric pressure.)

makkiz [27]

Given:

Water 4 m deep

Required:

Pressure at the bottom of the tank

Solution:

p2 – p1 = gh

p2 – p1 = p = gh

p = gh = 1000kg/m3 (9.8m/s2)(4m)

6 0

3 years ago

A. What is the frequency of blue light that has a wavelength of 450 nm?

erastova [34]

Answer

a)Given,

Wavelength of blue light = 450 n m = 450 x 10⁻⁹ m

Speed of light = 3 x 10⁸ m/s

we know,

$c = \nu \lambda$

$\nu = \dfrac{c}{\lambda}$

$\nu = \dfrac{3\times 10^8}{450\times 10^{-9}}$

$\nu = 6.67 \times 10^{14}\ Hz$

b)Given,

Wavelength of red light = 650 n m = 650 x 10⁻⁹ m

Speed of light = 3 x 10⁸ m/s

we know,

$c = \nu \lambda$

$\nu = \dfrac{c}{\lambda}$

$\nu = \dfrac{3\times 10^8}{650\times 10^{-9}}$

$\nu = 4.62 \times 10^{14}\ Hz$

3 0

3 years ago

a player passes a 0.600kg basketball down court for a fast break.the ball leaves the player hands with a speed of 8.30m/s and sl

Setler79 [48]

At the release point, the ball has kinetic energy,

$\dfrac12m{v_0}^2$

and at its maximum height it has potential and kinetic energy,

$-mgy+\dfrac12mv^2$

where $y$ is the maximum height attained above the release point.

The LoCoE tells us that we should have

$\dfrac12m{v_0}^2=-mgy+\dfrac12 mv^2$

$\implies v^2-{v_0}^2=-2gy$

$\implies\left(7.20\dfrac{\rm m}{\rm s}\right)^2-\left(8.30\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)y$

$\implies y=0.870\,\mathrm m$

7 0

2 years ago

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A horse runs 350 meters in 23 seconds. What was the average speed of the horse? Round your answer to a tenth ​

A horse runs 350 meters in 23 seconds. What was the average speed of the horse? Round your answer to a tenth