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Lubov Fominskaja [6]
2 years ago
11

PLEASE HELP!!! i’ll mark brainliest if you’re correct

Physics
1 answer:
dedylja [7]2 years ago
7 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

The net force acting on the block is ~

\qquad \sf  \dashrightarrow \: 10 + 5

\qquad \sf  \dashrightarrow \: 15 \:N

So, the Answer in the boxes will be ~

\boxed{ \sf15} \:  \boxed{ \sf N}

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A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
Please dont delete this question......
Juli2301 [7.4K]

Answer:

c

Explanation:

8 0
3 years ago
Read 2 more answers
Which of the following is true about the easy walk harness
Svetllana [295]
<span>the easy walk harness should be looser on the dog than other harnesses</span>
6 0
3 years ago
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An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0
Diano4ka-milaya [45]
The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = <span>µ x n x I where:
</span>µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = <span>µ </span><span>o = 4 pi x 10^-7 Tm/A
</span>I is the current passing (I = 4 amperes)
n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
B = 4 x 3.14 x 10^-7 x 5000 x 4 = 25.132 mT
5 0
3 years ago
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi
Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ *  6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

7 0
3 years ago
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