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igor_vitrenko [27]

1 year ago

8

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored, how high (in m) would

a projectile launched at this speed rise if projected straight up? m

1 answer:

motikmotik1 year ago

4 0

Then the maximum height of the catapult will be 69.78 meters.

<h3>What is kinematics?</h3>

The study of motion without considering the mass and the cause of the motion. The equation of motion is given below.

v = u + at

s = ut + (1/2)at²

v² = u² + 2as

A catapult can project a projectile at a speed as high as 37.0 m/s. (a) if air resistance can be ignored.

a = - 9.81 m/s²

u = 37 m/s

v = 0

s = h

Then the maximum height of the catapult will be given by the third equation.

v² = u² + 2as

0² = 37² - 2 × 9.81 × h

h = 37² / (2 × 9.81)

h = 69.78 meters

Then the maximum height of the catapult will be 69.78 meters.

More about the kinematics link is given below.

brainly.com/question/7590442

#SPJ1

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In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he

lana [24]

Answer:8.28 km

Explanation:

Given

First it drifts $45^{\circ}$ 2.5 km

$r_1=2.5cos45 i+2.5sin45 j$

Secondly it drifts $60^{\circ}$ 4.70 km

$r_{12}=4.7cos60 i-4.7sin60 j$

After that it drifted along east direction 5.1 km

$r_{23}=5.1 i$

After that it drifts $55^{\circ}$ 7.2 km

$r_{34}=-7.2cos55 i-7.2sin55 j$

After that it drifts $5^{\circ}$ 2.8 km

$r_{54}=-2.8cos5 i+2.8sin5 j$

$r_{5O}$ = $\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}$ + $\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}$

$r_{5O}=2.299\hat{i}-7.95\hat{j}$

$|r_{5O}|=8.28 km$

for direction

$tan\theta =\frac{7.95}{2.299}=3.4580$

$\theta =73.87^{\circ}$ south of east

7 0

3 years ago

Read 2 more answers

Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak

frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

$A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\$

Considering that the vector sum $A+B+C=0$ , then:

$|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0$

Then:

$V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0$

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

$V_{x}= A cos \alpha -B cos \beta=0 (1)\\V_{y}= A sin \alpha +B sin \beta - C=0 (2)$

Form Eq. (1):

$A=B \frac{cos \beta}{cos \alpha} (3)$

Replacing A in Eq. (2):

$(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1} (4)$

Replacing values of C, α and β in (4):

$B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1} \\B= -44.4$

Replacing value of B in (3)

$A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35$

5 0

3 years ago

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$

Where $c_p$ represent the specific heat for the water and this value from tables we know that $c_p =1 \frac{cal}{gr C}$ for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

8 0

3 years ago

Read 2 more answers

A toaster uses 6700 joules of energy in 45 seconds to toast to a piece of bread what is power of the oven

just olya [345]

P= w/t and W= Work
In this case, W= 6,700j, and T= 45 seconds
Power is the ratio of work per unit time. When you perform a work in a given span of time, the ratio of work performed with respect to time is Called Power.
si unit for Power is Watt (W)
so, P= 6,700/45
P= 148
Final answer is P=148

8 0

3 years ago

3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?

ivanzaharov [21]

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

6 0

3 years ago