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3 years ago

11

You push a box 3.5 m across the floor with a constant force of 12.4 N. How much work do you do on the box?

1 answer:

aliya0001 [1]3 years ago

3 0

Answer:

43.4 N (newton)

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Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27

Simora [160]

Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.

3 0

3 years ago

A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?

iren2701 [21]

Answer:

$K.E=29.403125J$

Explanation:

From the question we are told that

Mass $M=100$

Height $50-20=30m$

Generally the equation for velocity before impact is is is mathematically given by

$v=\sqrt{2gh}$

$v=\sqrt{2*9.8*30}$

$v=24.25$

Generally the equation for Kinetic Energy is is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*100*(24.25)^2\\$

$K.E=29403.125J$

$K.E=29.403125J$

8 0

3 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

7 0

3 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago

What is unique regarding the abdominal muscle when is compared to other muscle in the body?

marissa [1.9K]

Answer:a is the answer

8 0

3 years ago