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3 years ago

20. The hospital is....the police station. (with/on/by/between

Physics

1 answer:

aliya0001 [1]3 years ago

7 0

Answer:

Between

Explanation:

I think so because other doesn't make sense

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I need an answer for this plzz!!<br>number 2 <br>anybody can help ??

Anuta_ua [19.1K]

2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
$a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} }$
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
$2ax \: = \: {v}^{2} \: - \: {u}^{2}$
$x \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{2a} \: = \: \frac{{(25 \: \frac{m}{s})}^{2} \: - \: {(0 \: \frac{m}{s} )}^{2} }{2 \: \times \: 0.5 \: \frac{m}{ {s}^{2} } } \: = \: 625 \: m$
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.

8 0

3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 121.6 nanome

lina2011 [118]

Answer:

b) The star is moving away from us.

Explanation:

If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.

7 0

3 years ago

After clearing the bar on a high jump, you land softly on a giant mattress. landing on a mattress is more comfortable than landi

madam [21]

Answer:

The right approach is Option b (the force..................exert on you).

Explanation:

Even before you fall on something like a soft object, users eventually slow to a halt. You are still giving away all the downward momentum, but progressively although with small powers, you are doing so.
Although you can get injured by massive powers, this gradual displacement is a positive thing. And that is why you have a mattress you would like to settle on.

The other options given are not connected to the situation described. So, the solution here was the right one.

3 0

2 years ago

Which is the mode of the following data?<br>120, 125, 130, 125, 135

Aleks [24]

The mode in this case would be 125 because it occurs the most in the sequence of numbers.

8 0

3 years ago

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20. The hospital is....the police station. (with/on/by/between​

20. The hospital is....the police station. (with/on/by/between