Answer:
P = 1000000[Pa] = 1000 [kPa]
Explanation:
To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.
where:
P = pressure [Pa] (units of pascals)
F = force = 100 [N]
A = area = 100 [mm²]
But first we must convert the units from square millimeters to square meters.
![A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]](https://tex.z-dn.net/?f=A%3D100%5Bmm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B1000%5E%7B2%7Dmm%5E%7B2%7D%20%20%7D%20%3D0.0001%5Bm%5E%7B2%7D%20%5D)
Now replacing:
![P=100/0.0001\\P=1000000[Pa]](https://tex.z-dn.net/?f=P%3D100%2F0.0001%5C%5CP%3D1000000%5BPa%5D)
You fallow the steps to figure out the complete answer of the question.
Answer:K.E=449598.5j
Explanation:
Kinetic energy of a moving car=1/2mv^2
Where m is the mass of the car
And V is the velocity of the car
K.E=1/2 ×1300×26.3^2
K.E=449598.5j
Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).
Let's calculate the electric field generated by the first charge:
While the electric field generated by the second charge is
Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
Answer:
299.51 m/s
Explanation:
m = mass of the bullet = 45 g = 0.045 kg
M = mass of the block = 1.55 kg
v = muzzle speed of the bullet
V = speed of bullet-block combination after the collision
μ = Coefficient of friction between the block and the surface = 0.28
d = distance traveled by the block = 13 m
V' = final speed of the bullet-block combination = 0 m/s
acceleration of the bullet-block combination due to frictional force is given as
a = - μg
using the kinematics equation
V'² = V² + 2 a d
0² = V² + 2 (- μg) d
0 = V² - 2 (μg) d
0 = V² - 2 (0.28) (9.8) (13)
V = 8.45 m/s
Using conservation of momentum for collision between bullet and block
mv = (M + m) V
(0.045) v = (1.55 + 0.045) (8.45)
v = 299.51 m/s
