Answer:
Green and Black.
Explanation:
The white fur reflects all colors, so would appear green.
The black fur reflects no colors, so would still appear black.
wire has a resistance of 1.3 ohms. The crosssectional
area of this wire is
(1) 7.9 × 10^−8 m2 (3) 4.6 × 10^0 m2
(2) 1.1 × 10^−7 m2 (4) 1.3 × 10^7 m2
The cross-sectional area of the copper wire is which has 1.3 ohm at 20 degree Celsius during laboratory experiment by student is 7.9×10⁻⁸ ohm.
Resistance is obstacle for the current flow in the circuit. It is the measure of reverse act to current flow in through a material. It can be given as,
Here, (ρ) is the specific resistance (l) is the length of the wire and (A) is the cross-sectional area of the wire.
During a laboratory experiment, the student finds that at 20° Celsius, a 6.0-meter length of copper wire has a resistance of 1.3 ohms.
The value of resistivity of copper wire is 1.72×10⁻⁸ ohm-m. Put these values in the above formula as,
Thus, the cross-sectional area of the copper wire is which has 1.3 ohm at 20 degree Celsius during laboratory experiment by student is 7.9×10⁻⁸ ohm.
Learn more about the resistance here;
ρ = resistivity of copper at 20 °C = 1.72 x 10⁻⁸ ohm-m
R = resistance of the copper wire = 1.3 ohm
L = length of the copper wire = 6 meter
A = area of cross-section of the copper wire = ?
Resistivity of copper wire is given as
R = ρL/A
inserting the values in the above equation
1.3 = (1.72 x 10⁻⁸) (6)/A
A = 7.9 x 10⁻⁸ m²
Hence the correct choice is
(1) 7.9 x 10⁻⁸ m²
b.T1 is ..... M1g.
c. T3 is ..... m1g + M2g
d.T1 is ..... T2
e.The magnitude of the acceleration of M2 is ..... the magnitude of the acceleration on m1.
f. T1 + T2 is ..... T3
Answer:
a. center of mass acceleration supposed to be acceleration due to gravity, 9.81 m/s^2,
b. T1 = 9.81m1 N; c. T3 =9.81(M1+M2) N; d. T3-T1, e. (T3-T1)/M2; f. (M1+M2)T3/M3
Explanation:
In this frictionless, massless pulley system, the center of mass accelerates downward with an acceleration equal to the acceleration due to gravity. The tension in the string connected to mass M1 is equal to M1g, and the tension in the string connected to mass M2 is equal to m1g + M2g. The magnitudes of the accelerations of M1 and M2 are equal, and the sum of the tensions T1 and T2 is equal to the tension T3.
a. The center of mass accelerates: When considering the system as a whole, the acceleration of the center of mass is determined by the net external force acting on the system. In this case, the only external force is the force due to gravity. Therefore, the center of mass accelerates downward with an acceleration equal to g, the acceleration due to gravity.
b. T1 is equal to M1g: The tension in the string connected to mass M1 is equal to the weight of M1, which is given by the formula T1 = M1g.
c. T3 is equal to m1g + M2g: The tension in the string connected to mass M2 is equal to the sum of the weights of M1 and M2, which is given by the formula T3 = m1g + M2g.
d. T1 is equal to T2: Since the pulley is assumed to be frictionless and massless, the tension in the string connected to mass M1 is the same as the tension in the string connected to mass M2.
e. The magnitude of the acceleration of M2 is equal to the magnitude of the acceleration on M1: This is due to the constraint imposed by the tension in the string. Since the tension in the string connecting M1 and M2 is the same, their accelerations must also be the same.
f. T1 + T2 is equal to T3: The sum of the tensions T1 and T2 is equal to the tension T3, as the total force acting on mass M2 is equal to the sum of the individual tensions.
#SPJ12
The object of mass with 6.3 Kg will have an acceleration of 1.12 m/s² if a force of 7.1 Newtons is applied.
Force is an external agent acting on a body to change its motion or to deform it. There are various kinds of forces such as gravitational force, magnetic force, nuclear force etc.
Acceleration is a physical quantity that measures the rate of change in velocity of a body. Like force, it is a vector quantity. As per Newton's second law of motion, force is the product of mass and acceleration.
Hence, F = m a
given that, m = 6.3 Kg
F = 7.1 N.
then a = F/m
a = 7.1 N/6.3 kg
= 1.12 m/s²
Therefore, the acceleration of the object is 1.12 m/s².
Find more on acceleration :
#SPJ6
Answer:
F=ma
7.1=6.3×m
7.1÷6.3=m
m=1.126 kg