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Butoxors [25]
2 years ago
8

what is the maximum distance we can shoot a dart,from ground level provided our toy dart gun gives a maximum initial velocity of

2.7m/s and air resistance is negligible​
Physics
1 answer:
mestny [16]2 years ago
3 0

Answer:

R = v^2 sin 2 theta / g

The range provides the distance a projectile can travel

R(max) = v^2 / g    if theta = 45 deg

R = 2.7^2 / 9.8 = .74 m

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Protons and ____ have electric charge?
lidiya [134]
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5 0
3 years ago
Read 2 more answers
Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.0
AURORKA [14]

Answer:

N = 177843 sheets

Explanation:

We are given;

Mass;m = 0.0035 kg

Pressure; p = 101325 pa = 101325 N/m²

L = 0.279m

W = 0.216m

The weight of N sheets is N(mg)

Where;

m is the mass of one sheet

N is number of sheets

g is the acceleration due to gravity.

The pressure equals weight divided by the area on which the weight presses:

Thus,

p= F/A = Nmg/(L•W)

Therefore, making N the subject;

N = pLW/(mg)

N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)

N = 177843

5 0
3 years ago
Type the correct answer in each box. Use numerals instead of words.
ser-zykov [4K]

Answer:

6.23 newtons per second?

Explanation:

4 0
2 years ago
At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec
evablogger [386]

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

4 0
3 years ago
Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,
klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble=2\hat{i}-3\hat{k} N

Position vector=r=0.5\hat{j}-2\hat{k} m

a.We have to find the resulting torque on the pebble about origin.

Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

Torque=-k-1.5i-4jN-m

By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

\tau=(-2i+0.5j+k)\times (2i-3k)

\tau=-6j-k-1.5i+2j=-1.5i-4j-kN-m

6 0
3 years ago
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