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3 years ago

The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A:

Physics

2 answers:

FrozenT [24]3 years ago

6 0

the answer is point c

vichka [17]3 years ago

5 0

Answer:

The kinetic energy will be maximum at point C.

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At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at

noname [10]

$Given:\\m_b=100kg\\m_c=1600kg\\v_c=5.2 \frac{m}{s} \\p_b=p_c\\\\Find:\\v_b=?\\\\Solution:\\\\p_b=p_c\\\\p=mv\\\\m_bv_b=m_cv_c\Rightarrow v_b= \frac{m_cv_c}{m_b} \\\\v_b= \frac{1600kg\cdot5.2 \frac{m}{s} }{100kg} =83.2 \frac{m}{s}$

3 0

3 years ago

Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

nekit [7.7K]

Answer:

The magnitude of vector d is 16 and the angle with the x-axis is 270°

Explanation:

Operations With Vectors

Given two vectors in rectangular components:

$\vec a=(ax,ay)\ ,\ \vec b=(bx,by)$

The sum of the vectors is:

$\vec a+\vec b=(ax+bx,ay+by)$

The difference between the vectors is:

$\vec a-\vec b=(ax-bx,ay-by)$

The magnitude of $\vec a$ is:

$|\vec a|=\sqrt{ax^2+ay^2}$

The angle $\vec a$ makes with the horizontal positive direction is:

$\displaystyle \tan\theta=\frac{ay}{ax}\\$

The question provides the vectors:

$\vec a=(2,6)$

$\vec b=(2,22)$

$\vec d=\vec a-\vec b$

Calculate:

$\vec d=(2,6)-(2,22)=(0,-16)$

The magnitude of $\vec d$ is:

$|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16$

The angle is calculated by:

$\displaystyle \tan\theta=\frac{-16}{0}$

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0

3 years ago

The weight of a standard object defined as having a mass of exactly 2.9 kg is measured to be 28.449 n. in the same laboratory, a

PtichkaEL [24]

Mass of the object m = 2.9 kg
Force F1 = 28.449 N
F1 = m1 x a => a = F / m => 28.449 / 2.9 => a = 9.81, which is gravitational acceleration.
In the same lab, a = g = 9.81, second object F2 = 48.7N = m2 x a
m2 = F2 / a => 48.7 / 9.81 => m2 = 4.96 kg
Mass of the second object m2 = 4.96 kg

4 0

3 years ago

Read 2 more answers

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

erica [24]

Answer:

2N/cm

Step-by-step explanation:

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$