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nydimaria [60]
1 year ago
12

A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical c

omponent of its velocity after 4.0 s if air resistance is neglected?
Physics
1 answer:
FinnZ [79.3K]1 year ago
6 0
<h2>This vertical component of velocity is 35 m/s </h2>

Explanation:

The dart is project with 150 m/s from a point at an angle of 30⁰

The vertical component of velocity = 150 sin 30 = 75 m/s

Thus initial vertical velocity is = 75 m/s

The velocity after 4 s can be calculate by

v = u - g t

here u is the initial velocity and t is the time , g is the acceleration due to gravity .

Thus v = 75 - 10 x 4 = 35 m/s

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A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
adell [148]

Answer:

r = 0.114 m

Explanation:

To find the speed of the proton, from conservation of energy, we know that

KE = PE

Thus, we have;

(1/2)mv² = qV

Where;

V is potential difference = 1kv = 1000V

q is charge on proton which has a value of 1.6 x 10^(-19) C

m is mass of proton with a constant value of 1.67 x 10^(-27) kg

Let's make the velocity v the subject;

v =√(2qV/m)

v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))

v = 4.377 x 10^(5) m/s

Now to calculate the radius of the circular motion of charge we know that;

F = mv²/r = qvB

Thus, mv²/r = qvB

Divide both sides by v;

mv/r = qB

Thus, r = mv/qB

Value of B from question is 0.04T

Thus,

r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)

r = 0.114 m

r = 8.76 m

8 0
1 year ago
A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
2 years ago
A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi
Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

δ=1.18cm

6 0
2 years ago
Read 2 more answers
A submarine completed a 450 km training with an average speed of 50 km/h. For the first 180 km, it travelled at an average speed
Kryger [21]

Answer:

45km/hr

Explanation:

Total distance=450km

Total speed=50km/hr

Total time= distance/speed

=450/50

=9hrs

distance a=180km

speed a=60km/hr

Time a=180/60

=3hrs

Distance b=450-180=270km

Speed b=?

Time b=270/speed b

Total time=time a + time b

9=3+(270/speed b)

270/speed b =9-3

270/speed b =6

6*speed b =270

Speed b=270/6

Speed b=45km/hr

4 0
1 year ago
If a force of 26 N is exerted on two balls, one with a mass of 0.52 kg and the other with a mass of 0.78 kg, the ball with the m
gogolik [260]
False is the correct answer
6 0
2 years ago
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