Answer:
The Required horizontal force is 230.04N
Explanation:
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
where:
F_{Hn} is the required Force
u is the friction coefficient
m is the mass
g is gravitational acceleration=9.8m/s^2
Eq (1)
Now, weight increases by 42% and friction coefficient decreases by 19%
New weight=(1.42*m*g) and new friction coefficient=0.81u
Eq (2)
Divide Eq(2) and Eq (1)
The Required horizontal force is 230.04N
<span>1 C = 6.24150965(16)×10^18 electrons
31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs
</span><span>I hope this helps. </span>
<h2>Astronaut travels to different planets - Option 4 </h2>
If an astronaut travels to different planets, none of the planets will the astronaut’s weight be the same as on Earth. On jupiter, weight will be more than the weight on earth. For instance if an astronaut has 100kg on earth then he will have 252 kg on jupiter.
On Mars, weight will be less than the weight on the earth. For instance, if an astronaut has 68 kg on earth then he will has 26 kg on mars. On Mercury, weight of an astronaut will be less than the weight on earth. Example if he has 68 kg on earth then he will have 25.7kg on mercury.
Hence, none of these planets the weight of astronaut will be same as on earth.
Answer:
Explanation:
mass attached m = .14 kg
force constant k = 5N / m
displacement
= amplitude of oscillation
A = .03 m
A ) period of motion = 
= 2 x 3.14 
T = 1.05 s
B ) maximum speed of block = angular velocity x amplitude
= (2π /T) x A
= (2 x 3.14 x .03) / 1.05
= .1794 m / s
17.94 cm /s
C )
maximum acceleration = angular velocity² x amplitude
= (2π /T)² x A
= (2π /1.05)² x .03
= 1.073 m / s²
D )
position
S = A cos ωt , ω is angular velocity
S = .03 cos(2πt /T)
= .03 cos 5.98 t
v =ω A sin(2πt /T)
= 5.98 x .03 sin5.98t
= .1794 sin5.98t
acceleration = ω²A sin5.98t
= 1.073 sin5.98t
11 miles away from the direction she was running
