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Lilit [14]
2 years ago
4

Assume that at t1=0.25s the velocity of the object is v1=5.0cm/s. find the following velocities of the object: v0 at t0=0.0s, v2

at t2=0.50s , v3 at t3=0.75s, and v4 at t4=1.0s.
Physics
1 answer:
Zarrin [17]2 years ago
8 0
The equation for acceleration a is: a=v/t, where v is velocity and t is time. From there we can calculate the velocity: v=at, at any time if we know acceleration. If acceleration a is constant then:

a=v₁/t₁=5/0.25=20 cm/s²

Now we can find all velocities for the object:

v₀=at₀=20*0=0 cm/s
v₂=at₂=20*0.5= 10 cm/s
v₃=at₃=20*0.75=15 cm/s
v₄=at₄=20*1=20 cm/s 
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When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.
7 0
2 years ago
if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a
Whitepunk [10]

Answer:

r=0.41m

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

\tau=r\times F

Due to the definition of cross product, the magnitude of the torque is given by:

\tau=rFsin\theta

Where \theta is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when sin\theta is equal to one, solving for r:

r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m

7 0
2 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
2 years ago
What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?
erma4kov [3.2K]

<span>E = h x f </span>

<span>. . . then : </span>

<span>f = E / h </span>
<span>f = 4,41•10^-19 / 6,62•10^-34 </span>
<span>f = 6,66•10^14 Hz (s^-1) </span>


<span>b/ What is the wavelength of this light ? </span>
<span>- - - - - - - - - - - - - - - - - - - - - - - - - - - - </span>

<span>λ = c / f </span>
<span>λ = 3•10^8 / 6,66•10^14 </span>
<span>λ = 4,50•10^-7 m </span>
4 0
2 years ago
Read 2 more answers
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
2 years ago
Read 2 more answers
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