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2 years ago

What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?

2 answers:

5 0

49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma â‡’ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.

Lyrx [107]2 years ago

4 0

F=MA
F is force- we are going to solve for F
M is mass- quantity of matter that takes up an object
A is acceleration- change in velocity
F=MA
F=(0.25 kg)(196 m/s^2)
F= 49 N (newton-SI unit for force)
Hope this helps!

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3 years ago

Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max

yanalaym [24]

Answer:

$\theta=39.49^{\circ}$

Explanation:

Maximum height of the pumpkin, $H_{max}=9.99\ m$

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We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

$H_{max}=\dfrac{v^2\ sin^2\theta}{2g}$

On rearranging the above equation, to find the angle as :

$\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})$

$\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})$

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So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

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2 years ago

What is the smallest part of a chemical compound that has all the properties

mel-nik [20]

Answer:

D. A Molecule

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2 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

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2 years ago

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iVinArrow [24]

Path 2.

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