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castortr0y [4]
3 years ago
15

A newspaper is tossed at an angle of 35° above the horizontal. If it has a mass of 2.2 kg and a speed of 6 m/s, what is the magn

itude of the momentum in the horizontal direction?
A. 7.6 kg-m/s
B. 10.8 kg-m/s
C. 13.2 kg-m/s
D. 34.1 kg-m/s
Physics
1 answer:
tino4ka555 [31]3 years ago
3 0

(A). 7.6 kg-m/s is the answer

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Two planets have the same surface gravity, but planet b has twice the radius of planet
nevsk [136]
The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
m₂ = 4m

<em>Thus, the answer is D.</em>
8 0
3 years ago
If a vibrating string is made shorter (i.e., by holding your finger on it), what effect does
Ghella [55]
  <span>Pitch and frequency are more or less the same thing - high pitch = high frequency. 
The freqency of vibration of a string f = 1/length (L) so as length decreases the frequency increases.</span>
4 0
3 years ago
A box weighs 25N. How much mass does it have?
Rudiy27

Explanation:

If box weight 25N on ground

MA=F

M(10)=25

M=2.5Kg

3 0
3 years ago
Why heat capacity for polymer higher than metal?
jok3333 [9.3K]
Because polymers are covalently bonded material, and metals are metalicly bonded material. 

Covalent bonds do not let atoms exchange electrons like metalic bonds do.
4 0
3 years ago
A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th
Gala2k [10]

Answer:

F_a_v_g=7093333.33N*s

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}

Where:

m=mass\hspace{3}of\hspace{3}the\hspace{3}object

v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval

v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.

t_2=final\hspace{3}time

t_1=initial\hspace{3}time

Asumming v1=0 and t1=0:

F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s

8 0
4 years ago
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