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3 years ago

Is a hockey player moving at a constant velocity across frictionless ice, balance or unbalance

Physics

1 answer:

daser333 [38]3 years ago

6 0

Answer:

Balanced

Explanation:

An unbalanced force will cause a change in the velocity, either the speed or direction, thus if the object is moving at a constant velocity that indicates that there is no acceleration or change in velocity which means that the forces are balanced.

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If Abdul pedals his 20 kg bike with a force of 100 N, what will the bike’s acceleration be? The formulas above may be helpful in

Scorpion4ik [409]

Answer:

5 m/s/s

Explanation:

F = ma

100 = 20a

100/20 = a

a = 5 m/s/s

4 0

3 years ago

40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker a

lakkis [162]

Answer:

Explanation:

there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.

7 0

3 years ago

A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.

mr_godi [17]

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m$

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

5 0

3 years ago

According to Newton's first law, if an object is changing direction, what must be happening?

murzikaleks [220]

D. Unbalanced forces are acting on it.

hope this helped

6 0

3 years ago

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m

Gekata [30.6K]

Answer:

819.78 m

Explanation:

<u>Given:</u>

OA = range of initial position of the airplane from the point of observation = 375 m
OB = range of the final position of the airplane from the point of observation = 797 m
$\theta$ = angle of the initial position vector from the observation point = $43^\circ$
$\alpha$ = angle of the final position vector from the observation point = $123^\circ$
$\vec{AB}$ = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

$\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m$

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

$\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} = (-708.34\ \hat{i}+412.67\ \hat{j})\ m$

The magnitude of the displacement vector = $\sqrt{(-708.34)^2+(412.67)^2}\ m = 819.78\ m$

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.

8 0

3 years ago