Answer:
a) 23.51 m/s
b) 1.07 kg
Explanation:
Parameters given:
Kinetic energy, K = 295 J
Momentum, p = 25.1 kgm/s
a) The kinetic energy of a body is given as:
where m = mass of the body and v = speed of the body
We know that momentum is given as:
p = mv
Therefore:
K = 1/2 * pv
=> v = 2K / p
v = (2 * 295) / 25.1 = 23.51 m/s
The velocity of the body at that instant is 23.51 m/s.
b) Momentum is given as:
p = mv
=> m = p / v
m = 25.1 / 23.51 = 1.07 kg
The mass of the body at that instant is 1.07 kg
Answer: 0
Explanation:
75 newtons will push back canceling it out make it 0
Answer:
Explanation:
The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.
The strength of electric field is defined as the force experienced by the unit positive test charge.
E = F / q
Electric field strength is a vector quantity and it is measured in newton per coulomb.
Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.
The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.
Answer:
71.85 m/s
Explanation:
Given the following :
Length of skid marks left by jaguar (s) = 290 m
Skidding Acceleration (a) = - 8.90m/s²
Final velocity of jaguar (v) = 0
Speed of Jaguar before it Began to skid =?
Hence, initial speed of jaguar could be obtained using the formula :
v² = u² + 2as
Where
v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar
0² = u² + (2 × (-8.90) × 290)
0 = u² + (-5,162)
u² = 5162
Take the square root of both sides
u = √5162
u = 71.847 m/s
u = 71.85m/s
