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noname [10]
2 years ago
6

What is the purpose of using a wedge?

Physics
1 answer:
valentinak56 [21]2 years ago
4 0
Think of a wedge as something you put in between objects, so it is a separates objects
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Magnetism/ magnetic field ana magnetic forces
katovenus [111]

Answer:

Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.  

6 0
3 years ago
The Robinson projection map is considered very useful because
Vladimir79 [104]
Hello,
   The question states: <span>The Robinson projection map is considered very useful because...
The answer is \boxed{because \ most \ distances, \ sizes, \ and \ shapes \ are \ accurate.}

Hope this helped!

~FoodJunky
</span>
7 0
3 years ago
The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan
miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴  × 1000 m

r = 1.08  × 10⁷ m, G = 6.67  × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08  × 10⁷ )²/75( 6.67  × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0
3 years ago
A 4 cm diameter ball is located 40 cm from a point source and 80 cm from a wall. What is the size of the shadow on the wall?
Slav-nsk [51]
<span>11.823 cm There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall. The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm. Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So 82 + 1/21 * 2 = 82 + 2/21 = 82.0952381 Now we have the following dimensions with a circle replacing the ball in the original problem. Distance from wall to effective circle = 82.0952381 cm Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm Effective diameter of circle = 3.995462279 cm And because the geometry makes similar triangles, the following ratio applies. 3.995462279/41.9047619 = X/124 Now solve for X 3.995462279/41.9047619 = X/124 124*3.995462279/41.9047619 = X 495.4373226/41.9047619 = X 11.82293611 = X The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
7 0
3 years ago
How can a cyclist minimize friction as he or she rides?
tino4ka555 [31]
The answer would be B. All the other answers would add friction.
6 0
3 years ago
Read 2 more answers
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