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Lilit [14]
3 years ago
11

Help please!!!

Physics
1 answer:
exis [7]3 years ago
3 0

Answer:

5 years worth of work (aka all of the homework i currently have)

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The vertical component of the projectile motion of an object depends on which of these? initial velocity or angel of trajectory
SVEN [57.7K]
It depends on both of them.

In fact, the projectile begins its motion with an initial velocity of v_0 and an angle of \alpha. On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by
v_y (t) = v_0 sin \alpha + gt
and as it can be seen, this depends on both initial velocity and angle.
3 0
3 years ago
Do you think a moving skateboard has energy? Why or Why not? What about a skateboard that is not moving?
puteri [66]

Answer:

Yes

Explanation:

Because if you push it, the skateboard has kinetic/potential energy.

3 0
3 years ago
Read 2 more answers
The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7
stich3 [128]

Answer:

Electric field acting on the electron is  127500 N/C.

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Charge on electron, q=1.6\times 10^{-19}\ C

Initial speed of electron, u = 0

Final speed of electron, v=3\times 10^7\ m/s

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

F = ma

F=9.1\times 10^{-31}\times 2.25\times 10^{16}

F=2.04\times 10^{-14}\ N

Electric force is given by :

F = q E, E = electric field

E=\dfrac{F}{q}

E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
The force on the spring is F0 and it stores elastic potential energy PEs0. If the spring displacement is tripled to 3x0, determi
Dimas [21]

Answer:

Explanation:

Let initial extension in the spring= x₀

Force on the spring = F₀

Let spring constant = k

Fo = k x₀

Fn = 3k x₀

Fn /Fo = 3

PEs0 ( ORIGINAL)  =1/2 k x₀²

PEsn ( NEW)  =1/2 k (3x₀)²

PEsn / PEs0 = 9

7 0
3 years ago
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