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4 years ago

15

A distance-time graph indicates that an object 100 m in 4s and then remains at rest for 6s what is the average speed of the obje

ct

1 answer:

jonny [76]4 years ago

6 0

The average speed is 10 m/s.

Explanation:

The motion of the object is divided into two parts.

In the first part, the object covers a distance of

$d_1 = 100 m$

In a time interval of

$t_1 = 4 s$

In the second part, the object is a t rest, so it covers a distance of

$d_2 = 0$

In a time interval of

$t_2 = 6 s$

The average speed for the trip is given by the ratio between the total distance and the total time taken, therefore:

$s=\frac{d_1+d_2}{t_1+t_2}=\frac{100+0}{4+6}=10 m/s$

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What was anton van leeuwenhoek famous for

monitta

Answer:

He is known as the first microbiologist and also “the Father of Microbiology” because he was the first to observe bacteria underneath a microscope. He made many other significant discoveries in the field of biology and also made important changes to the microscope.

Explanation:

hope this helps. and if it did pls mark brainliest :)

3 0

3 years ago

An automobile steering wheel is shown. What is the ideal mechanical advantage? If the AMA is 8, what is the efficiency of the st

Mars2501 [29]

1. Ideal Mechanical Advantage (IMA): 9

Explanation:

For a wheel and axle system like the steering wheel, the IMA is given by:

$IMA=\frac{r_w}{r_a}$

where

$r_w$ is the radius of the wheel

$r_a$ is the radius of the axle

For the steering wheel of the problem, we see that $r_w = 18 cm$ and $r_a=2 cm$ , so the IMA is

$IMA=\frac{18 cm}{2 cm}=9$

2. Efficiency: 88.9%

Explanation:

The efficiency of a system is defined as the ratio between the AMA (actual mechanical advantage) and the IMA:

$\eta=\frac{AMA}{IMA}\cdot 100$

In this problem, AMA=8 and IMA=9, so the efficiency is

$\eta=\frac{8}{9}\cdot 100=88.9\%$

6 0

4 years ago

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

BaLLatris [955]

Answer:

= 4.3 × 10 ⁻¹⁴ m

Explanation:

The alpha particle will be deflected when its kinetic energy is equal to the potential energy

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

Kinetic energy of the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)

= 8.459 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³

= 4.3 × 10 ⁻¹⁴ m

3 0

4 years ago

When a can is crushed did it undergo a physical change

brilliants [131]

Because even though the object got crush and misshape it still has the same identity. the identity never change

4 0

4 years ago

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

$\triangle E = 12.79 J$

Explanation:

Sprinters' tendons stretch, $x_s = 43 mm = 0.043 m$

Non athletes' stretch, $x_n = 32 mm = 0.032 m$

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, $E_s = 0.5kx_s^2$

Maximum energy stored in the non athletes, $E_m = 0.5kx_n^2$

Difference in maximum stored energy between the sprinters and the non-athlethes:

$\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J$

4 0

4 years ago