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2 years ago

NEED HELP!! PHYSICS PROBLEM

Physics

1 answer:

garik1379 [7]2 years ago

7 0

Answer:

Explanation:

At Y, horizontal force is greater than downward force so E or L. However L seems to be more appropriate since horizontal force in L is less than E

At Z, horizontal force is less than downward force so D

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Find the center of the galaxy with Shapley method

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Answer:

Calculating Center

Shapley correctly determined the galactic center of the Milky Way to be located in the constellation of Sagittarius. He did this by mapping out a three-dimensional distribution of the globular clusters.

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2 years ago

A man pushes on piano with mass 170 kg; it slides at constant velocity down a ramp that is inclined at 20.0 ∘ above the horizont

nikdorinn [45]

Answer

given,

mass of the piano = 170 kg

angle of the inclination = 20°

moves with constant velocity hence acceleration = 0 m/s²

neglecting friction

so, force required to pull the piano

F = m g sin θ

F = 170 × 9.81 × sin 20°

F = 570.39 N

so, force required by the man to push the piano is F = 570.39 N

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3 years ago

A baseball accelerates downward at 9.8m/s. if the gravitational force acting on the baseball is 2.2n what is the baseballs mass

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3 years ago

a person is sitting on the last bench can see clearly see things written on book but cannot see them distinctly on board. what t

marysya [2.9K]

Answer:

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2 years ago

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

7 0

3 years ago