The equation of the car is given by the equation,
x(t) = 2.31 + 4.90t² - 0.10t⁶
If we are going to differentiate the equation in terms of x, we get the value for velocity.
dx/dt = 9.8t - 0.6t⁵
Calculate for the value of t when dx/dt = 0.
dx/dt = 0 = (9.8 - 0.6t⁴)(t)
The values of t from the equation is approximately equal to 0 and 2.
If we substitute these values to the equation for displacement,
(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31
(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51
Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.
Answer:
F = 326.7 N
Explanation:
given data
mass m = 200 kg
distance d = 2 m
length L = 12 m
solution
we know force exerted by the weight of the rock that is
W = m × g ..............1
W = 200 × 9.8
W = 1960 N
and
equilibrium the sum of the moment about that is
∑Mf = F(cos∅) L - W (cos∅) d = 0
here ∅ is very small so cos∅ L = L and cos∅ d = D
so F × L - W × d = 0 .................2
put here value
F × 12 - 1960 × 2 = 0
solve it we get
F = 326.7 N
Answer: i wasted one of your answer that sad srry
Explanation:
Answer:
yes driver exceed the car speed
Explanation:
given data
speed of car = 38 m/s
speed limit = 75.0 mi/h
to find out
Is the driver exceeding the speed limit
solution
we know car speed is 38 m/s and limit is 75 mi/hr
so for compare the speed limit we convert limit and make them same
as we know
1 m/s = 2.236 mi/hr
so
car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr
as this car speed is exceed the speed limit that is 75 mi/hr
yes driver exceed the car speed
solution:
the kinetic energy acquired by the electron when it is accelerated y an electric field is
k=5.25\times10^-16j
the kinetic energy acquired by the electron will ne proportional to the potential difference between the plates across which the electric field is applied.it is given by,
e\Delta v = k
here,\delta v is the potential difference between the plates, e is the charge on an electron and k is the kinetic energy.
reassange the above expression
\delta v = \frac{k}{e}
the potential difference between the plates will be
\delta v = \frac{k}{e}
sunstitute 5.25 \times 10^-16j for k and 1.6 \times10^-19c for e is the above equation
\delta v =\frac{5.25\times10^-16j}{1.60\times10^-19c}
=3.28\times10^3v(\frac{1kv}{1000v})
=3.28kv
since the electrons will e accelerated towards the plate at higher potential.
hance p;ate b will be at higher potential
