<span>The work done is 3.0 Nm.
We can us the equation Work = Force * Distance, where Force = 75.0 N, and distance is xf – xi = 3.00 cm - -1.00 cm = 4.00 cm. Convert centimeters to meters by moving the decimal place to the left by two places to get 0.04 m. Plug these values into the Work equation:
Work = Force * Distance
Work = 75.0 N * 0.04 m
Work = 3.0 Nm</span>
Answer:
H = 0.673
Explanation:
given,
side of cubical crate = 0.74
weight of the crate = 600 N
magnitude of force = 330 N
the Horizontal distance of its Center of mass
= 0.74/2
= 0.37
Let the required Height be H
By Balancing the Torques, we get
H x 330 N = 0.37 x 600
330 H = 222
H = 0.673
hence, the height above the floor where force is acting is equal to 0.673 m
Answer:
0.29 m/s due west.
Explanation:
According to newton's second law,
Net force acting on an object = mass×acceleration
From the question,
F+F₁+F₂ = ma................ Equation 1
Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.
Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg
substitute into equation 1
4080-680-1160 = 7660(a)
2240 = 7660a
Therefore,
a = 2440/7660
a = 0.29 m/s due west.
When two different air masses meet, a boundary is formed. the boundary between two air masses is called a front. weather at a front is usually cloudy and stormy. there at four different fronts: cold, warm, stationary, and occluded
The law of conservation of momentum tells us that momentum
is conserved, therefore total initial momentum should be equal to total final
momentum. In this case, we can expressed this mathematically as:
mA vA + mB vB = m v
where, m is the mass in kg, v is the velocity in m/s
since m is the total mass, m = mA + mB, we can write the
equation as:
mA vA + mB vB = (mA + mB) v
furthermore, car B was at a stop signal therefore vB = 0,
hence
mA vA + 0 = (mA + mB) v
1800 (vA) = (1800 + 1500) (7.1 m/s)
<span>vA = 13.02 m/s</span>