Answer:
3 seconds
Explanation:
To solve this problem, we have to find the acceleration of the car.
Acceleration is given as:
a = (v - u) / t
Where v = final velocity
u = initial velocity
t = time taken.
From the question,
u = 0 m/s
v = 31 mph = 13.86 m/s
t = 1.5 seconds
Acceleration, a, will be:
a (13.86 - 0) / 1.5
a = 13.86 / 1.5
a = 9.24 m/s²
We are told that this kind of car operates with constant force. This means it operates with constant acceleration (since force and acceleration are directly proportional)
Therefore, when the final velocity of the car is 62 mph (27.72 m/s), time taken is:
a = (v - u) / t
=> t = (v - u) / a
t = (27.72 - 0) / 9.24
t = 3 seconds
It will take the car 5 seconds to go from zero to 62 mph.
Answer:
2Mg + O2 = Reactant | 2MgO = Product
Explanation:
It is pretty simple, look:
The elements that are being added (or in the majority of the cases, the one who are in the left side) are always the Reactant.
The other side is the product of the reactants.
It is like a mathematic formula
We some the numbers to get the result, the numbers are the reactants and the result is the product.
(I hope you didn´t get confused)
Answer:
E = kQ / r^2 = 2.34*10^21 V/m
where k = 8.99*10^9 m/F
Q = 82 * 1.6*10^-19 C
r = 7.1*10^-15 m
Explanation:
14.2 FM = 2.34×10^21 V/m
The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
Learn more: brainly.com/question/8898885
Answer:
+%_×¥+₩ £kshnow sjsni niadk lplz kfkd skidk skkkim