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NISA [10]
2 years ago
15

What is physical quantity?....​

Physics
2 answers:
Art [367]2 years ago
6 0
Physical quantities are a characteristic or property of an object that can be measured or calculated from other measurements. Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as combinations of four fundamental units
Gnesinka [82]2 years ago
3 0
A physical quantity is any physical property of a material or system that can be quantified, that is, can be measured using numbers.
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A crumb of bread, of mass 0.056 kg, is pulled upon by ants from rival anthills. They exert the following forces: 0.06 N to the n
Valentin [98]

Answer:

a) 1.855m/s^2, 9.71\° to the east-north

b) 0.103N, 9.46° to the west-south

Explanation:

To find the acceleration of the system you can assume that  the forces are applied in a xy plane, where force toward north are directed in the +y direction, and forces to the east in the +x direction. BY taking into account the components of the acceleration for each axis you obtain the following systems of equations:

0.06N-0.06Nsin(45\°)=ma_y\\\\0.08N-0.02N+0.06cos(45\°)=ma_x

m: mass of the crumb of bread = 0.056kg

you simplify the equations an replace the values of the mass in order to obtain the acceleration components:

a_y=\frac{0.017N}{0.056kg}=0.313\frac{m}{s^2}\\\\a_x=\frac{0.102N}{0.056kg}=1.829\frac{m}{s^2}

\theta=tan^{-1}(\frac{0.313}{1.829})=9.71\°\\\\a=\sqrt{a_x^2+a_y^2}=\sqrt{(1.829)^2+(0.313)^2}\frac{m}{s^2}=1.855\frac{m}{s^2}

then, the acceleration of the system has a magnitude of 1.855m/s^2 and a direction of 9.71\° to the east-north

The fifth force must cancel both x an y components of the previous net force, that is:

0.06N-0.06Nsin(45\°)+F_y=0\\\\F_y=-0.017N\\\\0.08N-0.02N+0.06cos(45\°)+F_x=0\\\\F_x=-0.102N\\\\\phi=tan^{-1}(\frac{-0.017}{-0.102N})=9.46\°

F=\sqrt{(0.102)^2+(0.017)^2}N=0.103N

the, the force needed to reach the equilibrium has a magnitude of 0.103N and a direction of 9.46° to the west-south

4 0
3 years ago
Two general points to remember for recovering from skids are: keep your eyes pointed in the direction you want to go and you wil
Semmy [17]
The answer is do not break, the key avoiding skids is to always smoothly apply your brakes and accelerator and to turn slowly and smoothly. Reducing of the speed before oncoming turns and once driving in possibly hazardous circumstances such as wet, icy or snow covered roadways or on roadways with loose gravel.  
8 0
4 years ago
It's important to maintain muscle flexibility by stretching only after aerobic exercise.
lions [1.4K]

Answer: Most aerobic and strength training programs cause your muscles to contract and tighten. So stretching after you exercise helps optimize the range of motion about your joints and boosts circulation.

Hope this helps!

4 0
3 years ago
Suppose that you want to move a heavy box with mass 30.0 across a carpeted floor. You try pushing hard on one of the edges, but
timurjin [86]
The solution to the problem is as follows:
Normal force is m*g plus 240 N*sin30. 
<span>30 kg*9.8 m/s^2 + 240 N*sin30 = 414 N
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
7 0
3 years ago
To win the game, a place kicker must kick a
masya89 [10]

Answer:

0.57 m

Explanation:

First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is

v_x = u cos \theta = (15)(cos 51.7^{\circ})=9.30 m/s

And the distance to cover is

d = 19 m

So the time taken is

t=\frac{d}{v_x}=\frac{19}{9.30}=2.04 s

Now we want to find how high the ball is at that time. The initial vertical velocity is

u_y = u sin \theta = (15)(sin 51.7^{\circ})=11.77 m/s

So the vertical position of the ball at time t is

y(t) = u_y t - \frac{1}{2}gt^2

where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find

y=(11.77)(2.04)-\frac{1}{2}(9.8)(2.04)^2=3.62 m

The crossbar height is 3.05 m, so the difference is

\Delta h = 3.62 - 3.05 =0.57 m

So the ball passes 0.57 m above the crossbar.

8 0
3 years ago
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