3 years ago

A sprinter starts from rest and reaches a speed of 15 m/s in 4.25 s. Find his acceleration

Physics

1 answer:

kakasveta [241]3 years ago

5 0

Answer:

a=3.53 m/s^2

Explanation:

Vo=0 m/s (because he is not moving at the start)

V1=15 m/s

t= 4.25 s

a = (V1-Vo) / t = 15/4.25 = 3.53 m/s^2

You might be interested in

What is the gravitational force between two trucks, each with a masof 2.0x10^4 k

eduard

Answer:

$6.67\cdot 10^{-3} N$

Explanation:

The gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the two masses

r is the separation between the two masses

In this problem, we have

$m_1 = m_2 = 2.0 \cdot 10^4 kg$ the mass of each truck

r = 2.0 m is their separation

Substituting,

$F=(6.67\cdot 10^{-11}) \frac{(2.0\cdot 10^4 kg)^2}{(2.0 m)^2}=6.67\cdot 10^{-3} N$

7 0

4 years ago

What are the milestones of modern phyiscs?

nlexa [21]

Answer:

The articles appearing under "Milestones in Physics" will give an insight into special events or situations that have been decisive for the evolution of Physics

6 0

3 years ago

Read 2 more answers

An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre

irakobra [83]

Answer:

$T = -282.33^o C$

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

$P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)$

here we know that

$P_1 = 6.50 \times 10^4$

$P_o = 4.80 \times 10^4$

$T_1 = 100^o C$

$T_o = 0.01^o C$

now we will have

$P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)$

$P = 4.80 \times 10^4 + 170.02(T - 0.01)$

now if P = 0

then we will have

$0 = 4.80 \times 10^4 + 170.02(T - 0.01)$

$T = -282.33^o C$

7 0

3 years ago

PLEASE HELPPPPPPPPPPP

Oduvanchick [21]

It's easy raising the barbell.

3 0

3 years ago

Can anything block a magnet's force feild?

SVEN [57.7K]

Something that is not magnetic

7 0

4 years ago