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Naddik [55]
3 years ago
5

Which of these is a transparent object? A) window B) wax paper C) cell phone D) diamond ring

Physics
2 answers:
Ksivusya [100]3 years ago
8 0
I think it's A 
I hope that Help's
Sloan [31]3 years ago
3 0
The answer is A
Hope this helps:)
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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car’s moti
Nitella [24]
The car is staying still. either getting petrol or stopped at a light .

3 0
3 years ago
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An object has a mass of 2,000kg. what is its weight on earth? show your work​
artcher [175]

Answer:

F=m*g is the formula and the answer is 19,620 kg

Explanation:

Since the formula is F=m*g and Earth's gravity is 9.81 m/s^2 all you need to do is multiply 2,000 by 9.81

6 0
2 years ago
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A ball is kicked at a speed of 16m/s at 33° and it eventually returns to ground level further down field.
saw5 [17]

Hi there!

We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).

Remember to break the velocity into its vertical and horizontal components.

Thus:

0 = vi - at

0 = 16sin(33°) - 9.8(t)

9.8t = 16sin(33°)

t = .889 sec

Find the max height by plugging this time into the equation:

Δd = vit + 1/2at²

Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²

Solve:

Δd = 7.747 - 3.873 = 3.8744 m

4 0
3 years ago
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What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?
madam [21]

Answer:

there are 25 kg objective travelling at 2m/s to the right.

4 0
2 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
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