All categories

3 years ago

How far can a person run in 15 minutes if he runs at an average speed of 16 km/hr?

Physics

1 answer:

anygoal [31]3 years ago

3 0

Answer:

4km

Explanation:

15 minutes is 1/4 of an hour.

1/4 of 16 is 4.

You might be interested in

Se lanza una piedra de 3.00 N verticalmente hacia arriba desde el suelo. Se observa que, cuando está 15.0 m sobre el suelo, viaj

BARSIC [14]

Answer:

(a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

Explanation:

Given that,

Weight of stone = 3.00 N

Height = 15 m

Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

$W=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

Put the value into the formula

$-9.8\times15=\dfrac{1}{2}\times(v_{2}^2-v_{1}^2)$

$-2\times9.8\times15=25^2-v_{1}^2$

$-v_{1}^2=-300-25^2$

$v_{1}=\sqrt{925}$

$v_{1}=30.41\ m/s$

(b). We need to calculate the maximum height

Using work energy theorem

$[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

$mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

Here, $\dfrac{1}{2}mv_{2}^2$ =0

$-(mg)\times d=\dfrac{1}{2}mv_{1}^2$

$d=\dfrac{v_{1}^2}{2g}$

Put the value into the formula

$d=\dfrac{30.41^2}{2\times9.8}$

$d=47.18\ m$

Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

6 0

2 years ago

In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas

Sonja [21]

Answer:

The speed of disk is 1.98 $\frac{m}{s}$

Explanation:

Given:

Mass of $m = 0.099$ kg

Spring constant $k = 244 \frac{N}{m}$

Compression of spring $x = 4 \times 10^{-2}$ m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

$v = \sqrt{\frac{k x^{2} }{m} }$

$v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }$

$v = 1.98 \frac{m}{s}$

Therefore, the speed of disk is 1.98 $\frac{m}{s}$

8 0

2 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

2 years ago

Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1

Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0

2 years ago

The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of

jasenka [17]

1) Frequency: $3.29\cdot 10^{15}Hz$

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

$E=2.18 aJ=2.18\cdot 10^{-18} J$

The energy of a photon is given by

$E=hf$

where $h=6.63\cdot 10^{-34}Js$ is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

$f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz$

2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

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

where $c=3\cdot 10^8 m/s$ is the speed of light and f is the frequency. Substituting the frequency, we find

$\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm$

5 0

2 years ago