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3 years ago

14

A ball is thrown downward from the top of a 31.0 m tower with an initial speed of 10.0 m/s. Assuming negligible air resistance,

what is the speed of the ball just before hitting the ground

1 answer:

BARSIC [14]3 years ago

6 0

Answer:

$velocity \: = 26.833 {ms}^{ - 1}$

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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

3 0

3 years ago

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An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

3 years ago

Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each

jarptica [38.1K]

Answer with Explanation:

We are given that

$I_1=29 A$

$I_2=78 A$

$d=38 cm=\frac{38}{100}=0.38 m$

$1 m=100 cm$

a.Length of segment,l=20 m

Magnetic force ,F= $\frac{2\mu_0I_1I_2 l}{4\pi d}$

$\frac{\mu_0}{4\pi}=10^{-7}$

Substitute the values

$F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N$

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

4 0

3 years ago

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A wave hits a wall as shown. As the wave interacts with a wall, which kind of wave interaction is shown? absorption diffraction

dangina [55]

The answer is reflection.

The drawing is simple but illustrates the concept beautifully.

8 0

3 years ago

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The distance between 4 nodes (3 sections, 2 sections= wavelength) is 15.0cm. The frequency of the source is 10Hz. What's the spe

Natalija [7]

Answer:

the speed of the waves is 150 cm/s

Explanation:

Given;

frequency of the wave, f = 10 Hz = 10

distance between 4 nodes, L = 15.0 cm

The wavelength (λ) of the wave is calculated as follows;

Node to Node = λ/2

L = 2(Node to Node) = (4 Nodes) = 2 (λ/2) = λ

Thus, λ = L = 15.0 cm

The speed (v) of the wave is calculated as follows;

v = fλ

v = 10 Hz x 15.0 cm

v = 150 cm/s

Therefore, the speed of the waves is 150 cm/s

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3 years ago