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2 years ago

What is the horizontal component of a ball thrown at a 27 degree angle at 16 m/s?

Physics

1 answer:

Airida [17]2 years ago

6 0

Answer:

14.25 m/s

Explanation:

In this problem, we need to find the horizontal component of a ball thrown at a 27 degree angle at 16 m/s.

It can be given by :

$v_x=v\cos\theta\\\\=16\times \cos(27)\\\\=14.25\ m/s$

So, the horizontal component of the ball is 14.25 m/s.

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What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed

tekilochka [14]

Answer:

charge will be equal to $2.03\times 10^{-5}C$

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity $g=9.8m/sec^2$

Electric field E = 700 N/C

Electric force will be equal to $F=qE$ , here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

$q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C$

So charge will be equal to $2.03\times 10^{-5}C$

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2 years ago

An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring

BlackZzzverrR [31]

Answer:

a) $s = 0.603\,m$ , b) $s = 2.412\,m$

Explanation:

a) The system ice cube-spring is modelled by means of the Principle of Energy Conservation. Let assume that height at the bottom is zero:

$U_{g} = U_{k}$

$m\cdot g \cdot s\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x^{2}$

$s = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \sin \theta}$

$s = \frac{(25\,\frac{N}{m} )\cdot (0.1\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 0.603\,m$

b) The distance travelled by the ice cube is:

$s = \frac{(25\,\frac{N}{m} )\cdot (0.2\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 2.412\,m$

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2 years ago

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A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

$\theta=274rev$

Explanation:

From the question we are told that:

Angular velocity $\omega=510rpm$

Mass $m=40.kg$

Diameter d $75=>0.75m$

Off Time $t=40.0s$

Oscillation at Power off $N=210$

Generally the equation for Angular displacement is mathematically given by

$\theta_{\infty}=\frac{w+w_0}{t}t$

$w=\frac{2*\theta_{\infty}}{t}-w_0$

$w=\frac{28210}{40*(\frac{1}{60})}-510$

$w=120rpm$

Generally the equation for Time to come to rest is mathematically given by

$t=(\frac{\omega_0}{\omega_0-\omega})t$

$t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})$

$t=0.87min$

Therefore Angular displacement is

$\theta =(\frac{120+510}{2})0.87$

$\theta=274rev$

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2 years ago

Samples of different materials, A and B, have the same mass, but the sample

Effectus [21]

Answer:

B. The particles that make up material B have more mass than the

particles that make up material A.

Explanation:

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2 years ago

Tragically, the united states lost astronauts in accidents involvin

sweet-ann [11.9K]

The quest to put Americans on the moon before the Soviets do. During the Space Race, a couple of astronauts went up into space only to be tragically killed in an explosion.

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2 years ago

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