The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.
For this "type" of motion, displacement (Δx) can be determined by:
is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
= - 42
Displacement of the boat for t=6.0s interval is = - 42m, i.e., 42 m to the left.
Answer:
734.215N
Explanation:
First we calculate the angle that corresponds to a 5% slope using the Tan-1 function
then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)
Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is
Clock time of computer B is
Effective CPI of computer A is
Effective CPI of computer B is
CPU time of A is
CPU time of B is
Hence Computer A is Faster by
Computer A is 1.41 times faster than the Computer B
Answer:
<h2> 1.643*10⁻⁴cm</h2>Explanation:
In a single slit experiment, the distance on a screen from the centre point is expressed as y = where;
is the first two diffraction minima = 1
is light wavelength
d is the distance of diffraction pattern from the screen
a is the width of the slit
Given = 460-nm = 460*10⁻⁹m
d = 5.0mm = 5*10⁻³m
a = 1.4mm = 1.4*10⁻³m
Substituting this values into the formula above to get width of the central maximum y;
y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³
y = 2300*10⁻¹²/1.4*10⁻³
y = 1642.86*10⁻⁹
y = 1.643*10⁻⁶m
Converting the final value to cm,
since 100cm = 1m
x = 1.643*10⁻⁶m
x = 1.643*10⁻⁶ * 100
x = 1.643*10⁻⁴cm
Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is 1.643*10⁻⁴cm
Answer:
(a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
Explanation:
Given that,
Length = 50.00 cm
Time period = 2.50 s
Time period of pendulum is defined as the time for one complete cycle.
The period depends on the length of the pendulum.
Using formula of time period
Where, I = moment of inertia
We need to calculate the value of d
Using parallel theorem of moment of inertia
For a meter stick mass m , the rotational inertia about it's center of mass
Where, L = 1 m
Put the value into the formula of time period
Multiplying both sides by d
tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]
Put the value of T, L and g into the formula
The value of d is 0.056 cm and 1.496 cm.
(b). Given that,
L = 50-5 = 45 cm
We need to calculate the time period
Using formula of period
Put the value into the formula
Hence, (a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.