A :-) for this question , we should apply
F = ma
Given - F = 12 N
a = 0.20 m/s^2
Solution -
F = ma
12 = m x 0.20
m = 12 by 0.20
m = 60 kg
.:. The mass is 60 kg.
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Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;

E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
Answer:
1
Explanation:
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Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
= -k(T-Ta)
=
(T(t) -Ta)
=
= -k(T-Ta)
-ky
= -ky
T(t) -Ta = (To -Ta)
T(t) = Ta+ (To -Ta)
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) 
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) 
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min