Answer:
Points downward, and its magnitude is 9.8 m/s^2
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.
- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.
The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.
Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
Answer:
The height of the cliff is 90.60 meters.
Explanation:
It is given that,
Initial horizontal speed of the stone, u = 10 m/s
Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)
The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s
Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :
h = 90.60 meters
So, the height of the cliff is 90.60 meters. Hence, this is the required solution.
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge 
at a point 
distance away is given by
where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge .
= unit vector along .
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, 
are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
In the diagram, the ship send sound(?) waves to the water, to determine if there is anything there. If there is something like a sunken ship shown in the diagram, the waves return in a shorter time hence you can understand if theres something or now. This is the principle of radars and sonars.
