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givi [52]
2 years ago
5

Please help me.............................

Physics
1 answer:
Nadya [2.5K]2 years ago
7 0

Answer:

a

Explanation:

is a corect anser

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The slope of a line on a distance-time graph is
Nataly_w [17]

Answer: The slope of a line on a distance-time graph is- speed of the object.

The slope of a line on a graph refers to rate of change of variable that is presented on Y axis with respect to the variable that is presented on X axis.

For a distance time graph, distance is presented on Y axis and time on the X axis.

As we know that Speed= \frac{Distance}{time}

Therefore, the slope of a line on a distance-time graph represents speed of the object.

5 0
4 years ago
Read 2 more answers
Why is temperature constant during a phase change?
shutvik [7]
Because during a phase change nothing really happens. Before a phase change, energy and temperature will increase or decrease as well as after.
8 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
An electron has a velocity of 1.50 km/s (in the positive x direction) and an acceleration of 2.00 ✕ 1012 m/s2 (in the positive z
olga nikolaevna [1]

Answer:

see explanation

Explanation:

Given that,

velocity of 1.50 km/s = 1.50 × 10³m/s

acceleration of 2.00 ✕ 1012 m/s2

electric field has a magnitude of strength of 18.0 N/C

\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]

9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\

B_x = undetermined

B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T

B_z = 0T

8 0
4 years ago
A 1400 kg car moving +13.7 m/s makes an elastic collision with a 3200 kg truck, initially at rest. What is the velocity of the c
marishachu [46]

Answer:

-13.7m/s

Explanation:

If the car has a mass of 1400kg with a speed of +13.7m/s -> hits [email protected] rest and bounces with an elastic collision:

fm = -fm

Therefore, the speed the car will possess will be at a negative state.

= -13.7m/s

6 0
3 years ago
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