All categories

3 years ago

Salem walked 60 m in 5 seconds, his average speed is.

Physics

2 answers:

ZanzabumX [31]3 years ago

7 0

Answer:

Explanation:

Solution,

Distance(d)=60m

Time(t)=5sec

Now,

A.speed=d/t

A.speed=60/5=12m/s

dusya [7]3 years ago

3 0

Answer:

12m/s

Explanation:

60/5=12

You might be interested in

What do an earthquake wave and a light wave have in common?

rusak2 [61]

It’s c.) I think so at least

5 0

3 years ago

Read 2 more answers

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

- - - - - - - - - - - - -- - - - - - - - - - - - - -

TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

3 years ago

A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall

dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

$W_P=(171N)(8.80m)=1504.8J$

(b) The work don by the friction force is:

$W_f=F_fx=\mu N x=\mu Mg x$

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J$

$N=Mg=(46.0kg)(9,8m/s^2)=450.8N$

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

$W_N=0J$

(d) the same as before:

$W_g=0J$

8 0

3 years ago

What is the most likely reason that Mendeleev placed tellurium before iodine?

Verdich [7]

Mendeleev observed that tellurium has chemical properties like other elements in its group, and he did not know that neutrons cause the greater atomic mass is the most likely reason.

4 0

3 years ago

Read 2 more answers

Objects 1 and 2 attract each other with a gravitational force

snow_tiger [21]

Answer:

F' = 162 units

Explanation:

The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:

$F = \frac{Gm_{1}m_{2}}{r^{2}}\\\\$

where,

F = gravitational force = 18 units

G = Gravitational Constant

m₁ = mass of object 1

m₂ = mass of object 2

r = distance between objects

Therefore,

$18 = \frac{Gm_{1}m_{2}}{r^{2}}------ eqn (1)\\\\$

Now, if we change the value of distance to one-third of original value, then:

r' = r/3

$F' = \frac{Gm_{1}m_{2}}{(\frac{r}{3})^{2}}\\\\F' = (9)(\frac{Gm_{1}m_{2}}{r^{2}})$

using eqn (1):

F' = 9(18 units)

F' = 162 units

7 0

3 years ago