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3 years ago

A toy car accelerates at 3 m/s2 for 2 seconds. Its final velocity is 15 m/s, What is its initial velocity?

Physics

1 answer:

Talja [164]3 years ago

4 0

Answer:

can i have the toy car after your done asking questions with it

Explanation:

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Suppose a conducting rod is 52 cm long and slides on a pair of rails at 2.75 m/s. What is the strength of the magnetic field in

Eva8 [605]

Answer:

5.6 Tesla

Explanation:

L = 52 cm = 0.52 m

V = 2.75 m/s

e = 8 V

Let B be tha magnitude of magnetic field. Use the formula for the motional emf

e = B × V × L

B = e / V L

B = 8 / (2.75 × 0.52)

B = 5.6 Tesla

6 0

4 years ago

Two forces,

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at t = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at t = 10.4 s has x-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0

3 years ago

Which specific type of immunity is involved in antibody production?

Oksi-84 [34.3K]

The humoral response is mediated by B lymphocytes, which release antibodies specific to the infectious agent. The cell-mediated response involves the binding of cytotoxic T lymphocytes to foreign or infected cells, followed by the lysis of these cells.

6 0

3 years ago

You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the

Jet001 [13]

Answer:

Explanation:

Given

initial velocity $u=11.2\ m/s$

At maximum height velocity of ball is zero

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(0)^2 -(11.2)^2=2\times (-9.8)\times (s)$

$s=\frac{11.2^2}{2\times 9.8}$

$s=6.4$

time taken by the ball to reach the maximum height

$v=u+at$

$0=11.2-9.8\times t$

$t=\frac{11.2}{9.8}$

$t=1.142\ s$

At $t=2\ s$ height of ball is

$h=ut+\\frac{1}{2}at^2$

$h=11.2\times 2-\frac{1}{2}9.8\times (2)^2$

$h=22.4-19.6$

$h=2.8\ m$

i.e. ball is moving downward

height at $v=5\ m/s$

$v^2-u^2=2as$

$s=\frac{25-125.4}{2\times (-9.8)}$

$s=5.12\ m$

8 0

3 years ago

In a periodic table a set of properties repeats from?

bogdanovich [222]

Answer;

Row to row

In a periodic table a set of properties repeats from row to row

Explanation;

In a periodic table, a set of properties repeats from row to row. This is called the periodic law. In the periodic table, the rows are called periods and the columns are called groups.

The modern periodic table is organized by increasing atomic number. The atomic mass unit is based on 1/12th the mass of a carbon-12 atom.

The atomic mass of an element is the weighted average of the masses of an element’s isotopes.

8 0

3 years ago

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