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marusya05 [52]
3 years ago
8

Suppose you have a Frisbee with a mass of 0.10 kg. You first throw it with a force of 5N and then change the force you throw it

at to 20N. As your force increases, what is happening to the acceleration of your Frisbee?
Physics
1 answer:
natali 33 [55]3 years ago
8 0

Answer:

The acceleration increases.

Explanation:

From Newton's 2nd Law, we have \Sigma F=ma. We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.

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Dominik [7]
Can u show the whole question plz
3 0
3 years ago
A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

    M= I A

    M=16 \times 10^{-3} \times \pi \times r^2

   M=16 \times 10^{-3} \times \pi \times 0.3024^2

   M=4.59 \times 10^{-3}\ A m^2

b)  torque exerted in the loop

 \tau = M\ B

 \tau = 4.59 \times 10^{-3}\times 0.79

 \tau = 3.63 \times 10^{-3} N.m

8 0
3 years ago
With a mass of 109 kg, Baby Bird is the smallest monoplane ever
OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

  • m_1=61.4kg
  • m_2=109kg
  • v_1=1.84m/s
  • v_2=?

\\ \sf\longmapsto ∆P=P

\\ \sf\longmapsto m_1v_1=m_2v_2

\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}

\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}

\\ \sf\longmapsto v_2=112.976/109

\\ \sf\longmapsto v_2\approx 1.3m/s

4 0
3 years ago
If a t-shirt gun can fire t-shirts with an initial speed of 15 m/s, what is the maximum distance (along horizontal, flat ground)
kotegsom [21]

Answer:

h = 11.47 m

Explanation:

Initial speed pf the t-shirt gun is 15 m/s

We need to find the maximum distance covered by the t-shirt. It is based on the conservation of energy. The maximum distance covered is given by :

h=\dfrac{u^2}{2g}\\\\h=\dfrac{(15)^2}{2\times 9.8}\\\\h=11.47\ m

So, it will cover a distance of 11.47 m.

7 0
3 years ago
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Andru [333]

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

3 0
3 years ago
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