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lyudmila [28]
3 years ago
8

At what speed will a box be falling at a time t = 0.75 s after being dropped?

Physics
1 answer:
IRINA_888 [86]3 years ago
8 0

Explanation:

Initial speed(u)= 0 m/s (Ball is dropped)

time(t)= 0.75 s

acceleration(a)= 10 m/s² (gravity)

Final speed(v)= u+at

v=0+(10)× 0.75

v=7.5 m/s

Speed is 7.5 m/s

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Q1: At the highest point
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3 years ago
A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is
iris [78.8K]
(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the  coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
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\Phi = BA \cos \theta
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5 0
3 years ago
a hunter 412.5m from a cliff moves a distance x towards the cliff and fires a gun. he hears the echo from the cliff after 2.2sec
Inessa [10]

Answer: 49.5 m

Explanation:

The speed of sound s is given by a relation between the distance d and the time t:

s=\frac{d}{t} (1)

Where:

s=330 m/s is the speed of sound in air (taking into account this value may vary according to the medium the sound wave travels)

d=412.5 m-x since we are told th hunter was initially 412.5 meters from the cliff and then moves a distance x towards the cliff

t=\frac{2.2 s}{2}=1.1 s Since the time given as data (2.2 s) is the time it takes to the sound wave to travel from the hunter's gun and then go back to the position where the hunter is after being reflected by the cliff

Having this information clarified, let's isolate d and then find x:

d=st (2)

412.5 m-x=(330 m/s)(1.1 s) (3)

Finding x:

x=49.5 m This is the distance at which the hunter is from the cliff.

3 0
3 years ago
An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector
Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

v_{x} = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

t = \frac{X}{v_{x}}

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel  = 3.67 x 10⁻⁸ sec

v_{y} = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = v_{y} t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0
3 years ago
The diagram below shows the acceleration vs. time for a falling object. What was true about the object at 0.5 seconds?
Anvisha [2.4K]

Answer:

B. The object reached terminal velocity.

Explanation:

At t=0.5, the acceleration is 0, so the velocity is constant.  So the object has reached terminal velocity, and the net force is 0.

6 0
3 years ago
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