Answer:
417.5 kg
Explanation:
Since the gravitational for of attraction F, equals the weight of the object W when its close to the earth, F = GMm/R² = W = mg where m = mass of object and g = acceleration due to gravity close to the earth = 9.8 m/s².
g = GM/R² where g = acceleration due to gravity close to the earth, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.927 × 10²⁴ kg and R = radius of earth = 6.4 × 10⁶ m
So, g = GM/R²
= 6.67 × 10⁻¹¹ Nm²/kg² × 5.927 × 10²⁴ kg/(6.4 × 10⁶ m)²
= 39.53 × 10¹³ Nm²/kg ÷ 40.96 × 10¹² m²
= 9.65 N/kg
≅ 9.7 N/kg
= 9.7 m/s²
Since W = 4050 N and W = mg
m = W/g = 4050 N/9.7 m/s²
= 417.53 kg
≅ 417.5 kg
So, the mass of the object is 417.5 kg
What voltage is required to move 6A through 20?
Answer:
120V
Explanation:
Given parameters:
Current = 6A
Resistance = 20Ω
Unknown:
Voltage = ?
Solution:
According to ohms law;
V = IR
Where V is the voltage
I is the current
R is the resistance
Now, insert the parameters and solve;
V = 6 x 20 = 120V
what is the avarage speed during kick 1
the rubber covering over a wire acts as a / an ?
Answer:
Most electrical wire is covered in a rubber or plastic coating called insulation. The purpose of insulation covering the metal part of an electrical wire is to prevent accidental contact with other conductors of electricity, which might result in an unintentional electric current through those other conductors.
Explanation:
How much power does it take to do 1000 J of work in 8 seconds?
If you push an 5 N object 2 m and then push a 10 N object 2 m. Which is TRUE?
You do the same amount of work.
You do more work when you move the 10 N object.
You are more powerful when you move the 5 N object
O You use the same amount of force.
Answer:
Explanation:
You don't do the same amount of work. The work formula is F*d = W
W = work
F = force
d = the distance moved.
So you do more work when you move the 10N object because the Force (F) has doubled.
How much force would be needed to
push a 22 kg box across the floor with
an acceleration of 1.9 m/s^2? Assume
that there is a coefficient of friction of.
17 between the box and the floor.
Help me with this!!!!!!!it’s due in 10 minutes
There would be two forces acting on the box parallel to the floor, with a net force of
∑ F = p - f = m a
where p = magnitude of the push, f = mag. of friction, m = mass of the box, and a = acceleration. To find p, we first need f .
There are also only two forces acting on the box perpendicular to the floor, with net force
∑ F = n - w = 0
where n = mag. of normal force of the floor on the box and w = weight of the box. The net force is 0 because the box is only accelerating parallel to the floor.
w = m g, where g = 9.8 m/s² is the mag. of the acceleration due to gravity, so we can solve for n :
n = w = m g
n = (22 kg) (9.8 m/s²)
n = 215.6 N
The kinetic friction is proportional to the normal force by a factor of the given coefficient of friction, µ = 0.17, such that
f = µ n
f = 0.17 (215.6 N)
f = 36.652 N
Now solve for the required pushing force:
p - 36.652 N = (22 kg) (1.9 m/s²)
p ≈ 78 N
Answer:
0
Explanation: