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barxatty [35]
2 years ago
8

Which situation would create a field like the one shown here?

Physics
1 answer:
guapka [62]2 years ago
8 0
Huh? what situation??
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What is force applied per unit area
Marizza181 [45]

Answer:

Pressure is defined as force per unit area. The standard unit for pressure is the Pascal, which is a Newton per square meter.

P= A/F

3 0
3 years ago
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The vector product of vectors A⃗ and B⃗ has magnitude 12.0 m2 and is in the +z-direction.Vector A⃗ has magnitude 4.0 m and is in
g100num [7]

Answer:

θ=180°

Explanation:

The problem says that the vector product of A and B is in the +z-direction, and that the vector A is in the -x-direction. Since vector B has no x-component, and is perpendicular to the z-axis (as A and B are both perpendicular to their vector product), vector B has to be in the y-axis.

Using the right hand rule for vector product, we can test the two possible cases:

  • If vector B is in the +y-axis, the product AxB should be in the -z-axis. Since it is in the +z-axis, this is not correct.

  • If vector B is in the -y-axis, the product AxB should be in the +z-axis. This is the correct option.

Now, the problem says that the angle θ is measured from the +y-direction to the +z-direction. This means that the -y-direction has an angle of 180° (half turn).

8 0
3 years ago
How many significant figures are in the measurement 0.020 km?
Volgvan

Answer:3

Explanation:

4 0
3 years ago
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Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn
mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

5 0
3 years ago
A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to
dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m          

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So 1.41=2\times 3.14\sqrt{\frac{0.683}{K}}

0.2245=\sqrt{\frac{0.683}{K}}

Squaring both side

0.0504=\frac{0.4664}{K}

K=9.255N/m

So spring constant of the spring will be equal to 9.255 N /m

7 0
3 years ago
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