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3 years ago

11

Provides some background for this problem. a ball is thrown vertically upward, which is the positive direction. a little later i

t returns to its point of release. the ball is in the air for a total time of 7.70 s. what is its initial velocity? neglect air resistance.

1 answer:

julsineya [31]3 years ago

6 0

For an object thrown vertically upward, its time of flight is equal to the equation below:

t = 2v/g
where
t is the time of flight
v is the initial velocity
g is the acceleration due to gravity equal to 9.81 m/s²

Substituting the values,

7.07 = 2v/9.81
v = 34.68 m/s

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A current carrying wire is placed in a permanent magnetic field as shown in the diagram below. Determine the direction of the fo

USPshnik [31]

The direction of the magnetic force is to the right.

<h3>What is the magnetic field?</h3>

The magnetic field is the region in space where the influence of the magnet is felt. The magnetic force is always in the direction of the magnetic field.

We can see from the image, that the direction of the magnetic force is to the right.

Learn more about magnetic field:brainly.com/question/14848188

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7 0

2 years ago

A fisherman's scale stretches 3.7 cm when a 3.5 kg fish hangs from it. (a) what is the spring constant? n/m (b) what will be the

Karo-lina-s [1.5K]

(a) the weight of the fish is:
$W=mg=(3.5 kg)(9.81 m/s^2)=34.3 N$
and this is the force that stretches the spring by $x=3.7 cm=0.037 m$ . So, we can use Hook's law to find the constant of the spring:
$k= \frac{F}{x}= \frac{34.3 N}{0.037 m}=927.0 N/m$

(b) The fish is pulled down by 2.8 cm = 0.028 m more, so now the total stretch of the spring is
$x'=3.7 cm+2.8 cm=6.5 cm$
But this is also the amplitude of the new oscillation, because this is the maximum extension the spring can get, so A=6.5 cm.

The angular frequency of oscillation is given by:
$\omega= \sqrt{ \frac{k}{m} }= \sqrt{ \frac{927.0 N/m}{3.5 kg} }=16.3 Hz$
and so the frequency is given by
$f= \frac{\omega}{2 \pi} = \frac{16.3 Hz}{2 \pi}=2.6 Hz$

8 0

3 years ago

An object moves 2.5 m. This is an example of a _______.

makvit [3.9K]

The correct answer is A. Distance

Hope this helps

7 0

3 years ago

Read 2 more answers

Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet

Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times(5\times10^{-2})^2$

$A=7.85\times10^{-3}\ m^2$

We need to calculate the capacitor

Using formula of capacitor

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}$

$C=6.94\times10^{-12}\ F$

We need to calculate the resistance of the wire

Using formula of resistivity

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}$

$R=4.27\times10^{-3}\ \Omega$

We need to calculate the voltage

Using formula of charge

$q=CV$

$V=\dfrac{q}{C}$

Put the value into the formula

$V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}$

$V=1.801\times10^{3}\ V$

(a). We need to calculate the current

Using formula of current

$I=\dfrac{V}{R}$

$I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}$

$I=421779.85\ A$

$I=4.217\times10^{5}\ A$

(b). We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}$

$E=11.2\times10^{5}\ N/C$

The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

$E=\dfrac{1}{2}CV^2$

Put the value into the formula

$E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2$

$E=1.126\times10^{-5}\ J$

The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

Hence, (a).The maximum current in the wire is $4.217\times10^{5}\ A$ .

(b). The electric field in the wire is $11.2\times10^{5}\ N/C$ .

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is $1.126\times10^{-5}\ J$

8 0

3 years ago

If u accomplished 10,000 newton meters of work how much work did you do in units of joules

Amiraneli [1.4K]

1.0 joule= 1.0 newtons × 1.0 meter = 1.0 newton × meter

Work = 10 newtons × 5 meters = 50 newton × meter

3 0

3 years ago