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Leto [7]
3 years ago
14

Someone please help? :)

Physics
1 answer:
Gnoma [55]3 years ago
8 0

Answer:

(A) F_N - mg\cos\theta

Explanation:

The net force perpendicular to the surface of the incline is the sum of the gravity force component, which is mgcos(theta), and the reactionary normal force caused by the surface of the incline. The sum is F_N - mgcos(theta) and is usually 0 which is why the object is not moving perpendicularly to the surface of the incline.

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How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and
aev [14]

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h           ( g = 10 m/s²)

h= 46.66 m

4 0
3 years ago
A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th
pochemuha

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

  • <em>Initial momentum of the returner, </em>P_i_1<em> = 0 kgm/s</em>
  • <em>The initial momentum of the diving player, </em>P_i_2<em> = 130 kgm/s</em>

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0
2 years ago
particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

4 0
3 years ago
If R is the total resistance for a parallel circuit with two resistors of resistance r1 and r2, then . Find the resistance, r1,
Goshia [24]
For a parallel circuit with two resistors, the total resistance is calculated from the expression:

1/R = 1/R1 + 1/R2

We are given the total resistance, R, which is 20 ohms and R2 which is 75 ohms. We calculate R1 as follows:

1/20 = 1/R1 + 1/75
1/R1 = 11/300
R1 = 27.27 ohms
7 0
3 years ago
Read 2 more answers
Hi please may someone help me especially on the sketch part.
vaieri [72.5K]

Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.

We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s

we can use the formulae for acceleration to calculate the time taken/

(final - initial velocity)/timetaken=10

(30-0)/timetaken=10

timetaken =30/10=3 seconds

7 0
3 years ago
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