Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

$\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N$

$\rm F_v=F sin \theta \\\\ F_v=20 sin 50^0 \\\\ F_v=15.32 \ N$

The net vertical force is zero;

$\rm F_N=mg-Fsin50^0 \\\\ \rm F_N=8 \times 9.81 -15.32 \\\\ F_N=63.1 \ N \\\\$

From Newton's second law the net force as;

$\rm \sum F_{net}=ma \\\\ Fcos 60^0-f =ma \\\\ a=\frac{12.855-2.4}{8} \\\\a = 1.30 \ m/s^2$

Hence, the acceleration of the sled will be 1.30 m/s².

To learn more about the force refer to the link;

brainly.com/question/26115859

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5 0

2 years ago

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Winds are deflected to the (right/left) ___ as they move into a low pressure area in the Northern Hemisphere.

SashulF [63]

Winds are deflected to the right as they move into a low pressure area in the Northern Hemisphere.

<u>Explanation:</u>

Winds decide the motion of ocean currents which forms the surface waves in the Earth's atmosphere to maintain the pressure region. The motion of ocean currents is based on Coriolis force which states the direction of motion of an object in a rotating system.

In the case of Earth, the Coriolis force has an effect on the ocean currents which are deflected from maximum to minimum pressure region in a curved path. So the winds formed by the ocean currents will generally get deflected at the right as they move into a low pressure area at the Northern Hemisphere from the high pressure region.

8 0

3 years ago

We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,

White raven [17]

The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + $\frac{1}{2}$ g $t^{2}$