Answer:
Vf=60m/s
Vi=0 because it starts from rest
t=4.4s
a=?
a=Vf-Vi/t
a=60-0/4.4
a=60/4.4
a=13.63m/s^2
Answer:
13.636364 m/s2
Answer:
We conclude that it is a Bronsted-Lowry acid.
Explanation:
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According to Bronsted-Lowry an acid is a chemical species that is capable of yielding protons and a Bronsted-Lowry base is a chemical species capable of accepting protons.
In this case we see that sulfuric acid (H2SO4) loses a proton to become HSO4.
We conclude that it is a Bronsted-Lowry acid.
a. 10N
b. 100N
c. 200N
d. 98N
The smallest value of the force that will make the block not to slide down is 10 N.
We'll begin by calculating the normal reaction. This can be obtained as follow:
N = mg
N = 8 × 10
N = 80 N
Finally, we shall determine the frictional force.
F = μN
F = 0.4 × 80
F = 32 N
Since the frictional force is 32 N, therefore, a force lesser than the frictional force will make the blocknot to slide down.
From the options given above, only option A has a force that is lesserthan the frictional force.
Therefore, the correct answer to the question is Option A. 10 N
Learn more about frictional force:
The smallest value of the force that will not slide the 8.0 kg block down the wall is 31.36 N.
To determine the smallest value of the force such that the 8.0 kg block will not slide down the wall, we need to consider the static friction between the block and the wall. The formula for static friction is fs = μs * N, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which is mg = 8.0 kg * 9.8 m/s^2 = 78.4 N. Therefore, the smallest value of the force is equal to the maximum static friction force, which can be calculated as fs = 0.4 * 78.4 N = 31.36 N. So the correct answer is 31.36 N.
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Answer:
total stretch of the double-length spring will be 20 cm
Explanation:
given data
length x1 = 10 cm
mass = 1 kg
mass = double = 2 kg
to find out
the total stretch of the double-length spring will be
solution
we can say here spring constant is
k = mg ............1
k is spring constant and m is mass and g is acceleration due to gravity
so for in 1st case and 2nd case with 1 kg mass and 2 kg mass
kx1 = mg .........................2
and
kx2 = 2mg ........................3
x is length
so from equation 2 and 3
x2 = 20
so total stretch of the double-length spring will be 20 cm
B. potential energy.
C. thermal energy.
D. work.
Answer:
60
Explanation:
The volume of the water carried will be proportional to the cross-sectional area. The cross-section is a trapezoid with height 10sint, where t represents theta. The top of the trapezoid is 10+(2)(10cost), i.e. 10 + 20cost. The base of the trapezoid is 10.
Area of trapezoid = (average of bases) times (height)
= ([10 + (10 +20cost)] / 2 ) 10 sint
= (10 + 10cost)(10sint)
= 100sint + 100sintcost, call this A(t). You need to maximize. So differentiate and set equal to zero.
dA/dt = -100sin(t)^2+100cos(t)+100cos(t)^2 = 0, divide by 100:
-sin(t)^2 + cos(t) + cos(t)^2 = 0, replace sin^2 by 1-cos^2
2cos(t)^2 + cos(t) - 1 = 0, factor
(1+cos(t))(2cos(t)-1)=0, so
cos(t)=1, t=0 that give a min (zero area) not a max, or
cos(t) = 1/2, so t=60 degrees. This gives the max.