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2 years ago

An arrow is moving at 35 m/s and travels for 5 seconds. how far did the arrow travel?

Physics

1 answer:

pashok25 [27]2 years ago

8 0

The answer is 175 meters

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Please answer ASAP!!!

NARA [144]

The third (left hand corner) since the x and y are both negative.

4 0

2 years ago

A truck contains both hydraulic brakes (reliability 0.96) and mechanical brakes (reliability 0.99). What is the probability of s

Oksana_A [137]

Answer:

i) 0.9504

ii) 0.0452

Explanation:

Given data: reliability of hydraulic brakes= 0.96

reliability of mechanical brakes = 0.99

So the probability of stopping the truck = 0.96×0.99= 0.9504

At low speed

case: A works and B does not

= 0.96×(1-0.99) = 0.0096

case2 : B works and A does not

= 0.99×(1-0.96) = 0.0396

Therefore, probality of stopping = 0.0096+0.0396 = 0.0492

8 0

2 years ago

Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

Angular speed of the football spiral, $\omega=6.9\ rev/s=43.35\ rad/s$

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

$a=159.32\ m/s^2$

So, the centripetal acceleration of the laces on the football is $159.32\ m/s^2$ . Hence, this is the required solution.

6 0

2 years ago

when an object falls to the ground, only the object moves down but the earth's motion is not noticeable.why?

vlada-n [284]

Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.

7 0

2 years ago

The ground state energy of an electron in a one-dimensional trap with zero potential energy in the interior and infinite potenti

ElenaW [278]

Answer:

0.5 eV

Explanation:

$E_1$ = Initial potential energy = $2\ eV$

$E_2$ = Final potential energy

$L_1$ = Initial width

$L_2$ = Final width = $2L_1$

Energy of an electron in a one-dimensional trap is given by

$E=\dfrac{n^2h^2}{8mL^2}$

From the equation we get

$E\propto \dfrac{1}{L^2}$

So,

$\dfrac{E_1}{E_2}=\dfrac{L_2^2}{L_1^2}\\\Rightarrow E_2=\dfrac{E_1L_1^2}{L_2^2}\\\Rightarrow E_2=\dfrac{2L_1^2}{4L_1^2}\\\Rightarrow E_2=0.5\ eV$

The ground state energy will be 0.5 eV

6 0

2 years ago