The average force on the car=3857 N
Explanation:
mass of car= 1000 kg
initial velocity= Vi=0
Final velocity= Vf=27 m/s
time= t= 7 s
using the kinematic equation,
Vf= Vi + at
27=0+a (7)
a=3.857 m/s²
Now the force is given by F= m a
F=1000 (3.857)
F=3857 N
B
A prominence is a large, bright, gaseous feature extending outward from the Sun's surface, often in a loop shape. Prominences are anchored to the Sun's surface in the photosphere, and extend outwards into the Sun's corona.
Answer:
The final pressure will be 6.99 N/m^ 2
Explanation:
Initial volume of tire = 2.00 L
Initial pressure of the tire = 7.00 × 10 5 N/m 2
Initial temperature of tire 31 °C
P1V1 = P2V2
7.00 × 10 5 N/m 2 × V1 = 101325×100 → V1 = 1.45×10^(-5) m^3 = 1.45×10^(-2) L
Therefore new pressure = initial pressure - (pressure of released gas filling the entire 2 L)
pressure of released gas filling the entire 2 L = 2 L/(7.00 × 10^5 N/m^ 2 × 1.45×10^(-5) m^3) = 1.197 ×10^(-4) N/m^2
New pressure = 7.00 × 10^5 N/m^2 - 1.197 ×10^(-4) N/m^2
= 6.99 N/m^ 2
Answer:
E=1485.9N/C
Explanation: =
Electric field E=kQ/r²,
where k is a constant with a value of 8.99 x 10^9 N m²/C²
Q is the point charge
r is the distance between two point charges =0.44m
O=32*10^-9
Hence
E=(32*10^-9*8.99 x 10^9)/0.44²
E=32*8.99/0.1936
E=287.68/0.1936
E=1485.9N/C
Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
