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3 years ago

What would be 1/2 times 800?

Physics

1 answer:

IgorLugansk [536]3 years ago

7 0

Answer:

The answer is 400

Explanation:

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A 1425 kg truck driving at 13.0 m/s collides elastically with a stationary 1175 kg car. If the car is traveling 14.25 m/s just a

TEA [102]

Answer:

1.25 m/s

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(1425*13)+(1175*0)=(1425*v1f)+(1175*14.25)

18525+0=1425(v1f)+16743.75

1781.25=1425(v1f)

v1f=1.25 m/s

4 0

2 years ago

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago

Maggie completed a 10000-m race at an average speed of 160

Gala2k [10]

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

3 0

3 years ago

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.9 m/s. The car is a distanc

sammy [17]

If it takes $t$ seconds to reach the car, then the distance $d$ is $3.9t$ .

The bear's distance from the tourist's starting point is
$6t-23$

For maximum $d$ , we set the equations equal to each other:
$3.9t=6t-23$
$\Rightarrow -2.1t=-23$
$\Rightarrow t=\frac{23}{2.1}$
so the distance is
$d=3.9(\frac{23}{2.1})\approx42.741\ m$

6 0

3 years ago

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>

The time period of an oscillation refers to the time taken by an object to complete one oscillation.
It is the inverse of frequency of oscillation; denoted by "T".

Now,

$\sqrt{\frac{L}{g}}$ , where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
The period will increase as one climbs a very tall mountain because g will slightly decrease.
Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

#SPJ4

6 0

1 year ago