Ask question

Ask question

All categories

3 years ago

10

A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

tal distance that the dart can travel? Assume that the dart gun is fired at ground level.

1 answer:

In-s [12.5K]3 years ago

4 0

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

You might be interested in

Which of the following formulas is the correct equation for the law of Universal gravitation?

VladimirAG [237]

Answer:

F = GMm / r²

from Newton's law of gravitational attraction

8 0

3 years ago

What do we call air in the motion

Anna [14]

Answer:

carbon d

Explanation:

5 0

3 years ago

Read 2 more answers

In August 2011, the Juno spacecraft was launched from Earth with the mission of orbiting Jupiter in 2016. The closest distance b

Vikentia [17]

(a) 8927 mi/h

In order to calculate the average speed, we need to convert the time (t=5.0 y) into hours first. In 1 year, we have 365 days, each day consisting of 24 hours, so the time taken is:

$t=(5.0 y)(365 d/y)(24 h/d)=43,800 h$

The distance covered by the spacecraft is

$d=391 mil. mi = 391\cdot 10^6 mi$

Therefore, the average speed is just the ratio between the distance covered and the time taken:

$v=\frac{d}{t}=\frac{391\cdot 10^6 mi}{43,800 h}=8,927 mi/h$

(b) 35 minutes (2097 seconds)

The transmitted signals (which is a radio wave, which is an electromagnetic wave) travels back to the Earth at the speed of light:

$c=3.0\cdot 10^8 m/s$

Since 1 miles = 1609 metres, the distance covered by the signal is

$d=391\cdot 10^6 mi \cdot (1609 m/mi)=6.29\cdot 10^{11} m$

So, the time taken by the signal will be

$t=\frac{d}{v}=\frac{6.29\cdot 10^{11} m}{3.0\cdot 10^8 m/s}=2097 s$

And since 1 minute = 60 sec, the time taken is

$t=2097 s \cdot \frac{1}{60 s/min}\sim 35 min$

7 0

3 years ago

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

Marina86 [1]

Answer: $\alpha =10.66 rad/s^2$

Explanation:

Given

mass of disk $m=5 kg$

diameter of disc $d=30 cm$

Force applied $F=4 N$

Now this force will Produce a torque of magnitude

$T=F\cdot r$

$T=4\dot 0.15$

$T=0.6 N-m$

And Torque is given Product of moment of inertia and angular acceleration $(\alpha )$

$T=I\cdot \alpha$

Moment of inertia for Disc $I= \frac{Mr^2}{2}$

$I=0.05625 kg-m^2$

$0.6=0.05625\cdot \alpha$

$\alpha =10.66 rad/s^2$

3 0

3 years ago

A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s

sladkih [1.3K]

I believe the answer is option C 1.8 kg•m/s to the east

6 0

3 years ago