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4 years ago

James rode his bike 0.65 hours and traveled 8.45 km. What was his speed?

1 answer:

5 0

Speed = (distance covered) / (time to cover the distance)

= ( 8.45 km) / (0.65 hr)

= (8.45 / 0.65) km/hr

= 13 km/hr

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If a ball goes at 20 miles per hour what's the velocity

stellarik [79]

The velocity is

(20 miles per hour) in the direction (that the ball is traveling this instant) .

If the speed OR direction of the ball changes, then the velocity has changed.

8 0

4 years ago

A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of

son4ous [18]

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

7 0

3 years ago

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind th

Sedbober [7]

Answer:

c = 4,444.44

Explanation:

You have the following expression for the acceleration of the projectile:

$a=cs$ (1)

s: distance to the ground of the projectile

To find the value of the constant c you use the following formula:

$v^2=v_o^2+2a \Delta s$ (2)

vo: initial velocity = 0 m/s

v: final speed = 200 m/s

Δs: distance traveled by the projectile = 3m - 1.5m = 1.5m

You replace the expression (1) into the expression (2):

$v^2=2(cs)\Delta s$

You do the constant c in the last equation, then you replace the values of v, s and Δs:

$c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44$

6 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

How much time does it take the cheetah to travel 500 meters, if its average speed is 70 meters per second?

Anettt [7]

l=500m

speed=70m/s

t=?

t=500/70=7.14s

7 0

3 years ago